2 Standard formulation of non-barotropic magnetohydrodynamics

The standard set of equations solved for non-barotropic magnetohydrodynamics are given below:

(2.1) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial \boldsymbol{B}}{\partial t}=\boldsymbol{{\rm\nabla}}\times (\boldsymbol{v}\times \boldsymbol{B}), & \displaystyle\end{eqnarray}$$

(2.2) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }\boldsymbol{B}=0, & \displaystyle\end{eqnarray}$$

(2.3) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial {\it\rho}}{\partial t}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}\boldsymbol{v})=0, & \displaystyle\end{eqnarray}$$

(2.4) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\rho}\frac{\text{d}\boldsymbol{v}}{\text{d}t}={\it\rho}\left(\frac{\partial \boldsymbol{v}}{\partial t}+(\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}})\boldsymbol{v}\right)=-\boldsymbol{{\rm\nabla}}p({\it\rho},s)+\frac{(\boldsymbol{{\rm\nabla}}\times \boldsymbol{B})\times \boldsymbol{B}}{4{\rm\pi}}, & \displaystyle\end{eqnarray}$$

(2.5) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}s}{\text{d}t}=0. & \displaystyle\end{eqnarray}$$

The following notations are utilized: $\partial /\partial t$ is the temporal derivative, $\text{d}/\text{d}t$ is the temporal material derivative and $\boldsymbol{{\rm\nabla}}$ has its standard meaning in vector calculus. $\boldsymbol{B}$ is the magnetic field vector, $\boldsymbol{v}$ is the velocity field vector, ${\it\rho}$ is the fluid density and $s$ is the specific entropy. Finally $p({\it\rho},s)$ is the pressure which depends on the density and entropy (the non-barotropic case).

The justification for those equations and the conditions under which they apply can be found in standard books on magnetohydrodynamics (see for example Sturrock (Reference Sturrock1994)). The above applies to a collision-dominated plasma in local thermodynamic equilibrium. Such conditions are seldom satisfied by physical plasmas, certainly not in astrophysics or in fusion-relevant magnetic confinement experiments. Nevertheless it is believed that the fastest macroscopic instabilities in those systems obey the above equations (Yahalom Reference Yahalom2013a ), while instabilities associated with viscous or finite conductivity terms are slower. It should be noted that due to a theorem by Bateman (Reference Bateman1931) every physical system can be described by a variational principle (including viscous plasma) the trick is to find an elegant variational principle usually depending on a small number of variational variables. The current work will discuss only ideal magnetohydrodynamics while viscous magnetohydrodynamics will be left for future endeavours.

Equation (2.1) describes the fact that the magnetic field lines are moving with the fluid elements (‘frozen’ magnetic field lines), (2.2) describes the fact that the magnetic field is solenoidal, (2.3) describes the conservation of mass and (2.4) is the Euler equation for a fluid in which both pressure and Lorentz magnetic forces apply. The term:

(2.6) $$\begin{eqnarray}\boldsymbol{J}=\frac{\boldsymbol{{\rm\nabla}}\times \boldsymbol{B}}{4{\rm\pi}},\end{eqnarray}$$

is the electric current density which is not connected to any mass flow. Equation (2.5) describes the fact that heat is not created (zero viscosity, zero resistivity) in ideal non-barotropic magnetohydrodynamics and is not conducted, thus only convection occurs. The number of independent variables for which one needs to solve is eight ( $\boldsymbol{v},\boldsymbol{B},{\it\rho},s$ ) and the number of equations (2.1), (2.3)–(2.5) is also eight. Notice that (2.2) is a condition on the initial $\boldsymbol{B}$ field and is satisfied automatically for any other time due to (2.1).

3 Variational principle of non-barotropic magnetohydrodynamics

In the following section we will generalize the approach of Yahalom & Lynden-Bell (Reference Yahalom and Lynden-Bell2008) for the non-barotropic case. Consider the action:

(3.1) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle A\equiv \int {\mathcal{L}}\,\text{d}^{3}x\,\text{d}t,\\ \displaystyle {\mathcal{L}}\equiv {\mathcal{L}}_{1}+{\mathcal{L}}_{2},\\ \displaystyle {\mathcal{L}}_{1}\equiv {\it\rho}\left({\displaystyle \frac{1}{2}}\boldsymbol{v}^{2}-{\it\varepsilon}({\it\rho},s)\right)+{\displaystyle \frac{\boldsymbol{B}^{2}}{8{\rm\pi}}},\\ \displaystyle {\mathcal{L}}_{2}\equiv {\it\nu}\left[{\displaystyle \frac{\partial {\it\rho}}{\partial t}}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}\boldsymbol{v})\right]-{\it\rho}{\it\alpha}{\displaystyle \frac{\text{d}{\it\chi}}{\text{d}t}}-{\it\rho}{\it\beta}{\displaystyle \frac{\text{d}{\it\eta}}{\text{d}t}}-{\it\rho}{\it\sigma}\frac{\text{d}s}{\text{d}t}-\frac{\boldsymbol{B}}{4{\rm\pi}}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}\times \boldsymbol{{\rm\nabla}}{\it\eta}.\end{array}\right\} & \displaystyle\end{eqnarray}$$

In the above ${\it\varepsilon}$ is the specific internal energy (internal energy per unit of mass). The reader is reminded of the following thermodynamic relations which will become useful later:

(3.2) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle \text{d}{\it\varepsilon}=T\,\text{d}s-P\,\text{d}{\displaystyle \frac{1}{{\it\rho}}}=T\,\text{d}s+{\displaystyle \frac{P}{{\it\rho}^{2}}}\,\text{d}{\it\rho}\\ \displaystyle {\displaystyle \frac{\partial {\it\varepsilon}}{\partial s}}=T,\quad {\displaystyle \frac{\partial {\it\varepsilon}}{\partial {\it\rho}}}=\frac{P}{{\it\rho}^{2}}\\ \displaystyle w={\it\varepsilon}+{\displaystyle \frac{P}{{\it\rho}}}={\it\varepsilon}+{\displaystyle \frac{\partial {\it\varepsilon}}{\partial {\it\rho}}}{\it\rho}={\displaystyle \frac{\partial ({\it\rho}{\it\varepsilon})}{\partial {\it\rho}}}\\ \displaystyle \text{d}w=\text{d}{\it\varepsilon}+\text{d}\left({\displaystyle \frac{P}{{\it\rho}}}\right)=T\,\text{d}s+{\displaystyle \frac{1}{{\it\rho}}}\,\text{d}P\end{array}\right\} & \displaystyle\end{eqnarray}$$

in the above $T$ is the temperature and $w$ is the specific enthalpy. Obviously ${\it\nu},{\it\alpha},{\it\beta},{\it\sigma}$ are Lagrange multipliers which were inserted in such a way that the variational principle will yield the following equations:

(3.3) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle {\displaystyle \frac{\partial {\it\rho}}{\partial t}}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}\boldsymbol{v})=0,\\ \displaystyle {\it\rho}{\displaystyle \frac{\text{d}{\it\chi}}{\text{d}t}}=0,\\ \displaystyle {\it\rho}\frac{\text{d}{\it\eta}}{\text{d}t}=0,\\ \displaystyle {\it\rho}\frac{\text{d}s}{\text{d}t}=0.\end{array}\right\} & \displaystyle\end{eqnarray}$$

It is not assumed that ${\it\nu},{\it\alpha},{\it\beta},{\it\sigma}$ are single valued. Provided ${\it\rho}$ is not null those are just the continuity equation (2.3), entropy conservation and the conditions that Sakurai’s functions are comoving. Taking the variational derivative with respect to $\boldsymbol{B}$ we see that

(3.4) $$\begin{eqnarray}\boldsymbol{B}=\hat{\boldsymbol{B}}\equiv \boldsymbol{{\rm\nabla}}{\it\chi}\times \boldsymbol{{\rm\nabla}}{\it\eta}.\end{eqnarray}$$

Hence $\boldsymbol{B}$ is in Sakurai’s form and satisfies (2.2). It can be easily shown that provided that $\boldsymbol{B}$ is in the form given in (3.4), and (3.3) are satisfied, then also (2.1) is satisfied.

For the time being we have showed that all the equations of non-barotropic magnetohydrodynamics can be obtained from the above variational principle except Euler’s equations. We will now show that Euler’s equations can be derived from the above variational principle as well. Let us take an arbitrary variational derivative of the above action with respect to $\boldsymbol{v}$ , this will result in:

(3.5) $$\begin{eqnarray}\displaystyle {\it\delta}_{\boldsymbol{v}}A & = & \displaystyle \int \text{d}t\left\{\int \text{d}^{3}x\,\text{d}t{\it\rho}{\it\delta}\boldsymbol{v}\boldsymbol{\cdot }[\boldsymbol{v}-\boldsymbol{{\rm\nabla}}{\it\nu}-{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}-{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}-{\it\sigma}\boldsymbol{{\rm\nabla}}s]\right.\nonumber\\ \displaystyle & & \displaystyle +\left.\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }{\it\delta}\boldsymbol{v}{\it\rho}{\it\nu}+\int \text{d}{\it\bf\Sigma}\boldsymbol{\cdot }{\it\delta}\boldsymbol{v}{\it\rho}[{\it\nu}]\right\}.\end{eqnarray}$$

The integral $\oint \,\text{d}\boldsymbol{S}\boldsymbol{\cdot }{\it\delta}\boldsymbol{v}{\it\rho}{\it\nu}$ vanishes in many physical scenarios. In the case of astrophysical flows this integral will vanish since ${\it\rho}=0$ on the flow boundary, in the case of a fluid contained in a vessel no-flux boundary conditions ${\it\delta}\boldsymbol{v}\boldsymbol{\cdot }\hat{\boldsymbol{n}}=0$ are induced ( $\hat{\boldsymbol{n}}$ is a unit vector normal to the boundary). The surface integral $\int \,\text{d}{\it\bf\Sigma}$ on the cut of ${\it\nu}$ vanishes in the case that ${\it\nu}$ is single valued and $[{\it\nu}]=0$ as is the case for some flow topologies. In the case that ${\it\nu}$ is not single valued only a Kutta-type velocity perturbation (Yahalom et al. Reference Yahalom, Pinhasi and Kopylenko2005) in which the velocity perturbation is parallel to the cut will cause the cut integral to vanish. An arbitrary velocity perturbation on the cut will indicate that ${\it\rho}=0$ on this surface which is contradictory to the fact that a cut surface is to some degree arbitrary as is the case for the zero line of an azimuthal angle. We will show later that the ‘cut’ surface is comoving with the flow hence it may become quite complicated. This uneasy situation may be somewhat be less restrictive when the flow has some symmetry properties.

Provided that the surface integrals do vanish and that ${\it\delta}_{\boldsymbol{v}}A=0$ for an arbitrary velocity perturbation we see that $\boldsymbol{v}$ must have the following form:

(3.6) $$\begin{eqnarray}\boldsymbol{v}=\hat{\boldsymbol{v}}\equiv \boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s.\end{eqnarray}$$

The above equation is reminiscent of Clebsch representation in non-magnetic fluids (Clebsch Reference Clebsch1857, Reference Clebsch1859). Let us now take the variational derivative with respect to the density ${\it\rho}$ we obtain:

(3.7) $$\begin{eqnarray}\displaystyle {\it\delta}_{{\it\rho}}A & = & \displaystyle \int \text{d}^{3}x\,\text{d}t{\it\delta}{\it\rho}\left[\frac{1}{2}\boldsymbol{v}^{2}-w-\frac{\partial {\it\nu}}{\partial t}-\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\nu}\right]\nonumber\\ \displaystyle & & \displaystyle +\int \text{d}t\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }\boldsymbol{v}{\it\delta}{\it\rho}{\it\nu}+\int \text{d}t\int \text{d}{\it\bf\Sigma}\boldsymbol{\cdot }\boldsymbol{v}{\it\delta}{\it\rho}[{\it\nu}]+\int \text{d}^{3}x{\it\nu}{\it\delta}{\it\rho}|_{t_{0}}^{t_{1}}.\end{eqnarray}$$

In which $w=\partial ({\it\varepsilon}{\it\rho})/\partial {\it\rho}$ is the specific enthalpy. Hence provided that $\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }\boldsymbol{v}{\it\delta}{\it\rho}{\it\nu}$ vanishes on the boundary of the domain and $\int \,\text{d}{\it\bf\Sigma}\boldsymbol{\cdot }\boldsymbol{v}{\it\delta}{\it\rho}[{\it\nu}]$ vanishes on the cut of ${\it\nu}$ in the case that ${\it\nu}$ is not single valuedFootnote ¹ and the appropriate term vanish at initial and final times the following equation must be satisfied:

(3.8) $$\begin{eqnarray}\frac{\text{d}{\it\nu}}{\text{d}t}=\frac{1}{2}\boldsymbol{v}^{2}-w.\end{eqnarray}$$

Since the right-hand side of the above equation is single valued as it is made of physical quantities, we conclude that:

(3.9) $$\begin{eqnarray}\frac{\text{d}[{\it\nu}]}{\text{d}t}=0.\end{eqnarray}$$

Hence the cut value is comoving with the flow and thus the cut surface may become arbitrarily complicated. This uneasy situation may be somewhat be less restrictive when the flow has some symmetry properties.

Finally we have to calculate the variation with respect to both ${\it\chi}$ and ${\it\eta}$ this will lead us to the following results:

(3.10) $$\begin{eqnarray}\displaystyle {\it\delta}_{{\it\chi}}A & = & \displaystyle \!\int \text{d}^{3}x\,\text{d}t{\it\delta}{\it\chi}\left[\frac{\partial ({\it\rho}{\it\alpha})}{\partial t}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}{\it\alpha}\boldsymbol{v})-\boldsymbol{{\rm\nabla}}{\it\eta}\boldsymbol{\cdot }\boldsymbol{J}\right]+\!\int \text{d}t\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }\left[\frac{\boldsymbol{B}}{4{\rm\pi}}\times \boldsymbol{{\rm\nabla}}{\it\eta}-\boldsymbol{v}{\it\rho}{\it\alpha}\right]{\it\delta}{\it\chi}\nonumber\\ \displaystyle & & \displaystyle +\int \text{d}t\int \text{d}{\it\bf\Sigma}\boldsymbol{\cdot }\left[\frac{\boldsymbol{B}}{4{\rm\pi}}\times \boldsymbol{{\rm\nabla}}{\it\eta}-\boldsymbol{v}{\it\rho}{\it\alpha}\right][{\it\delta}{\it\chi}]-\int \text{d}^{3}x{\it\rho}{\it\alpha}{\it\delta}{\it\chi}|_{t_{0}}^{t_{1}},\end{eqnarray}$$

(3.11) $$\begin{eqnarray}\displaystyle {\it\delta}_{{\it\eta}}A & = & \displaystyle \!\int \text{d}^{3}x\,\text{d}t{\it\delta}{\it\eta}\left[\frac{\partial ({\it\rho}{\it\beta})}{\partial t}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}{\it\beta}\boldsymbol{v})+\boldsymbol{{\rm\nabla}}{\it\chi}\boldsymbol{\cdot }\boldsymbol{J}\right]+\!\int \text{d}t\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }\left[\boldsymbol{{\rm\nabla}}{\it\chi}\times \frac{\boldsymbol{B}}{4{\rm\pi}}-\boldsymbol{v}{\it\rho}{\it\beta}\right]{\it\delta}{\it\eta}\nonumber\\ \displaystyle & & \displaystyle +\int \text{d}t\int \text{d}{\it\bf\Sigma}\boldsymbol{\cdot }\left[\boldsymbol{{\rm\nabla}}{\it\chi}\times \frac{\boldsymbol{B}}{4{\rm\pi}}-\boldsymbol{v}{\it\rho}{\it\beta}\right][{\it\delta}{\it\eta}]-\int \text{d}^{3}x{\it\rho}{\it\beta}{\it\delta}{\it\eta}|_{t_{0}}^{t_{1}}.\end{eqnarray}$$

Provided that the correct temporal and boundary conditions are met with respect to the variations ${\it\delta}{\it\chi}$ and ${\it\delta}{\it\eta}$ on the domain boundary and on the cuts in the case that some (or all) of the relevant functions are non-single valued, we obtain the following set of equations:

(3.12a,b ) $$\begin{eqnarray}\frac{\text{d}{\it\alpha}}{\text{d}t}=\frac{\boldsymbol{{\rm\nabla}}{\it\eta}\boldsymbol{\cdot }\boldsymbol{J}}{{\it\rho}},\quad \frac{\text{d}{\it\beta}}{\text{d}t}=-\frac{\boldsymbol{{\rm\nabla}}{\it\chi}\boldsymbol{\cdot }\boldsymbol{J}}{{\it\rho}},\end{eqnarray}$$

in which the continuity equation (2.3) was taken into account. By correct temporal conditions we mean that both ${\it\delta}{\it\eta}$ and ${\it\delta}{\it\chi}$ vanish at initial and final times. As for boundary conditions which are sufficient to make the boundary term vanish one can consider the case that the boundary is at infinity and both $\boldsymbol{B}$ and ${\it\rho}$ vanish. Another possibility is that the boundary is impermeable and perfectly conducting. A sufficient condition for the integral over the ‘cuts’ to vanish is to use variations ${\it\delta}{\it\eta}$ and ${\it\delta}{\it\chi}$ which are single valued. It can be shown that ${\it\chi}$ can always be taken to be single valued, hence taking ${\it\delta}{\it\chi}$ to be single valued is no restriction at all. In some topologies ${\it\eta}$ is not single valued and in those cases a single valued restriction on ${\it\delta}{\it\eta}$ is sufficient to make the cut term null.

Finally we take a variational derivative with respect to the entropy $s$ :

(3.13) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}_{s}A=\int \text{d}^{3}x\,\text{d}t{\it\delta}s\left[\frac{\partial ({\it\rho}{\it\sigma})}{\partial t}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}{\it\sigma}\boldsymbol{v})-{\it\rho}T\right]+\int \text{d}t\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }{\it\rho}{\it\sigma}\boldsymbol{v}{\it\delta}s-\int \text{d}^{3}x{\it\rho}{\it\sigma}{\it\delta}s|_{t_{0}}^{t_{1}},\hspace{-6.0pt} & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

in which the temperature is $T=\partial {\it\varepsilon}/\partial s$ . We notice that according to (3.6) ${\it\sigma}$ is single valued and hence no cuts are needed. Taking into account the continuity equation (2.3) we obtain for locations in which the density ${\it\rho}$ is not null the result:

(3.14) $$\begin{eqnarray}\frac{\text{d}{\it\sigma}}{\text{d}t}=T,\end{eqnarray}$$

provided that ${\it\delta}_{s}A$ vanished for an arbitrary ${\it\delta}s$ .

4 Euler’s equations

We shall now show that a velocity field given by (3.6), such that the equations for ${\it\alpha},{\it\beta},{\it\chi},{\it\eta},{\it\nu},{\it\sigma},s$ satisfy the corresponding equations (3.3), (3.8), (3.12), (3.14) must satisfy Euler’s equations. Let us calculate the material derivative of  $\boldsymbol{v}$ :

(4.1) $$\begin{eqnarray}\frac{\text{d}\boldsymbol{v}}{\text{d}t}=\frac{\text{d}\boldsymbol{{\rm\nabla}}{\it\nu}}{\text{d}t}+\frac{\text{d}{\it\alpha}}{\text{d}t}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\alpha}\frac{\text{d}\boldsymbol{{\rm\nabla}}{\it\chi}}{\text{d}t}+\frac{\text{d}{\it\beta}}{\text{d}t}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\beta}\frac{\text{d}\boldsymbol{{\rm\nabla}}{\it\eta}}{\text{d}t}+\frac{\text{d}{\it\sigma}}{\text{d}t}\boldsymbol{{\rm\nabla}}s+{\it\sigma}\frac{\text{d}\boldsymbol{{\rm\nabla}}s}{\text{d}t}.\end{eqnarray}$$

It can be easily shown that:

(4.2) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle {\displaystyle \frac{\text{d}\boldsymbol{{\rm\nabla}}{\it\nu}}{\text{d}t}}=\boldsymbol{{\rm\nabla}}{\displaystyle \frac{\text{d}{\it\nu}}{\text{d}t}}-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial {\it\nu}}{\partial x_{k}}}=\boldsymbol{{\rm\nabla}}\left({\displaystyle \frac{1}{2}}\boldsymbol{v}^{2}-w\right)-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial {\it\nu}}{\partial x_{k}}},\\ \displaystyle {\displaystyle \frac{\text{d}\boldsymbol{{\rm\nabla}}{\it\eta}}{\text{d}t}}=\boldsymbol{{\rm\nabla}}{\displaystyle \frac{\text{d}{\it\eta}}{\text{d}t}}-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial {\it\eta}}{\partial x_{k}}}=-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial {\it\eta}}{\partial x_{k}}},\\ \displaystyle {\displaystyle \frac{\text{d}\boldsymbol{{\rm\nabla}}{\it\chi}}{\text{d}t}}=\boldsymbol{{\rm\nabla}}{\displaystyle \frac{\text{d}{\it\chi}}{\text{d}t}}-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial {\it\chi}}{\partial x_{k}}}=-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial {\it\chi}}{\partial x_{k}}},\\ \displaystyle {\displaystyle \frac{\text{d}\boldsymbol{{\rm\nabla}}s}{\text{d}t}}=\boldsymbol{{\rm\nabla}}{\displaystyle \frac{\text{d}s}{\text{d}t}}-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial s}{\partial x_{k}}}=-\boldsymbol{{\rm\nabla}}v_{k}{\displaystyle \frac{\partial s}{\partial x_{k}}}.\end{array}\right\} & \displaystyle\end{eqnarray}$$

In which $x_{k}$ is a Cartesian coordinate and a summation convention is assumed. Inserting the result from (4.2), (3.3) into (4.1) yields:

(4.3) $$\begin{eqnarray}\displaystyle \frac{\text{d}\boldsymbol{v}}{\text{d}t} & = & \displaystyle -\boldsymbol{{\rm\nabla}}v_{k}\left(\frac{\partial {\it\nu}}{\partial x_{k}}+{\it\alpha}\frac{\partial {\it\chi}}{\partial x_{k}}+{\it\beta}\frac{\partial {\it\eta}}{\partial x_{k}}+{\it\sigma}\frac{\partial s}{\partial x_{k}}\right)+\boldsymbol{{\rm\nabla}}\left(\frac{1}{2}\boldsymbol{v}^{2}-w\right)+T\boldsymbol{{\rm\nabla}}s\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{{\it\rho}}((\boldsymbol{{\rm\nabla}}{\it\eta}\boldsymbol{\cdot }\boldsymbol{J})\boldsymbol{{\rm\nabla}}{\it\chi}-(\boldsymbol{{\rm\nabla}}{\it\chi}\boldsymbol{\cdot }\boldsymbol{J})\boldsymbol{{\rm\nabla}}{\it\eta})\nonumber\\ \displaystyle & = & \displaystyle -\boldsymbol{{\rm\nabla}}v_{k}v_{k}+\boldsymbol{{\rm\nabla}}\left(\frac{1}{2}\boldsymbol{v}^{2}-w\right)+T\boldsymbol{{\rm\nabla}}s+\frac{1}{{\it\rho}}\boldsymbol{J}\times (\boldsymbol{{\rm\nabla}}{\it\chi}\times \boldsymbol{{\rm\nabla}}{\it\eta})\nonumber\\ \displaystyle & = & \displaystyle -\frac{\boldsymbol{{\rm\nabla}}p}{{\it\rho}}+\frac{1}{{\it\rho}}\boldsymbol{J}\times \boldsymbol{B}.\end{eqnarray}$$

In which we have used both (3.6) and (3.4) in the above derivation. This of course proves that the non-barotropic Euler equations can be derived from the action given in (3.1) and hence all the equations of non-barotropic magnetohydrodynamics can be derived from the above action without restricting the variations in any way except on the relevant boundaries and cuts.

5 Simplified action

The reader of this paper might argue here that the paper is misleading. The author has declared that he is going to present a simplified action for non-barotropic magnetohydrodynamics instead he added six more functions ${\it\alpha},{\it\beta},{\it\chi},{\it\eta},{\it\nu},{\it\sigma}$ to the standard set $\boldsymbol{B},\boldsymbol{v},{\it\rho},s$ . In the following I will show that this is not so and the action given in (3.1) in a form suitable for a pedagogic presentation can indeed be simplified. It is easy to show that the Lagrangian density appearing in (3.1) can be written in the form:

(5.1) $$\begin{eqnarray}\displaystyle {\mathcal{L}} & = & \displaystyle -{\it\rho}\left[\frac{\partial {\it\nu}}{\partial t}+{\it\alpha}\frac{\partial {\it\chi}}{\partial t}+{\it\beta}\frac{\partial {\it\eta}}{\partial t}+{\it\sigma}\frac{\partial s}{\partial t}+{\it\varepsilon}({\it\rho},s)\right]+\frac{1}{2}{\it\rho}[(\boldsymbol{v}-\hat{\boldsymbol{v}})^{2}-(\hat{\boldsymbol{v}})^{2}]\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{8{\rm\pi}}[(\boldsymbol{B}-\hat{\boldsymbol{B}})^{2}-(\hat{\boldsymbol{B}})^{2}]+\frac{\partial ({\it\nu}{\it\rho})}{\partial t}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\nu}{\it\rho}\boldsymbol{v}).\end{eqnarray}$$

In which $\hat{\boldsymbol{v}}$ is a shorthand notation for $\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s$ (see (3.6)) and $\hat{\boldsymbol{B}}$ is a shorthand notation for $\boldsymbol{{\rm\nabla}}{\it\chi}\times \boldsymbol{{\rm\nabla}}{\it\eta}$ (see (3.4)). Thus ${\mathcal{L}}$ has four contributions:

(5.2) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle {\mathcal{L}}=\hat{{\mathcal{L}}}+{\mathcal{L}}_{\boldsymbol{v}}+{\mathcal{L}}_{\boldsymbol{B}}+{\mathcal{L}}_{boundary},\\ \displaystyle \hat{{\mathcal{L}}}\equiv -{\it\rho}\left[{\displaystyle \frac{\partial {\it\nu}}{\partial t}}+{\it\alpha}{\displaystyle \frac{\partial {\it\chi}}{\partial t}}+{\it\beta}{\displaystyle \frac{\partial {\it\eta}}{\partial t}}+{\it\sigma}{\displaystyle \frac{\partial s}{\partial t}}+{\it\varepsilon}({\it\rho},s)+{\displaystyle \frac{1}{2}}(\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s)^{2}\right]\\ \displaystyle -\,{\displaystyle \frac{1}{8{\rm\pi}}}(\boldsymbol{{\rm\nabla}}{\it\chi}\times \boldsymbol{{\rm\nabla}}{\it\eta})^{2}\\ \displaystyle {\mathcal{L}}_{\boldsymbol{v}}\equiv {\textstyle \frac{1}{2}}{\it\rho}(\boldsymbol{v}-\hat{\boldsymbol{v}})^{2},\\ \displaystyle {\mathcal{L}}_{\boldsymbol{B}}\equiv {\displaystyle \frac{1}{8{\rm\pi}}}(\boldsymbol{B}-\hat{\boldsymbol{B}})^{2},\\ \displaystyle {\mathcal{L}}_{boundary}\equiv {\displaystyle \frac{\partial ({\it\nu}{\it\rho})}{\partial t}}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\nu}{\it\rho}\boldsymbol{v}).\end{array}\right\} & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

The only term containing $\boldsymbol{v}$ isFootnote ² ${\mathcal{L}}_{\boldsymbol{v}}$ , it can easily be seen that this term will lead, after we nullify the variational derivative with respect to $\boldsymbol{v}$ , to (3.6) but will otherwise have no contribution to other variational derivatives. Similarly the only term containing $\boldsymbol{B}$ is ${\mathcal{L}}_{\boldsymbol{B}}$ and it can easily be seen that this term will lead, after we nullify the variational derivative, to (3.4) but will have no contribution to other variational derivatives. Also note that the term ${\mathcal{L}}_{boundary}$ contains only complete partial derivatives and thus cannot contribute to the equations, although it can change the boundary conditions. Hence we see that (3.3), (3.8), (3.12) and (3.14) can be derived using the Lagrangian density:

(5.3) $$\begin{eqnarray}\displaystyle & & \displaystyle \hat{{\mathcal{L}}}[{\it\alpha},{\it\beta},{\it\chi},{\it\eta},{\it\nu},{\it\rho},{\it\sigma},s]\nonumber\\ \displaystyle & & \displaystyle \quad =-{\it\rho}\left[\frac{\partial {\it\nu}}{\partial t}+{\it\alpha}\frac{\partial {\it\chi}}{\partial t}+{\it\beta}\frac{\partial {\it\eta}}{\partial t}+{\it\sigma}\frac{\partial s}{\partial t}+{\it\varepsilon}({\it\rho},s)+\frac{1}{2}(\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s)^{2}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1}{8{\rm\pi}}(\boldsymbol{{\rm\nabla}}{\it\chi}\times \boldsymbol{{\rm\nabla}}{\it\eta})^{2}\end{eqnarray}$$

in which $\hat{\boldsymbol{v}}$ replaces $\boldsymbol{v}$ and $\hat{\boldsymbol{B}}$ replace $\boldsymbol{B}$ in the relevant equations. Furthermore, after integrating the eight equations (3.3), (3.8), (3.12), (3.14) we can insert the potentials ${\it\alpha},{\it\beta},{\it\chi},{\it\eta},{\it\nu},{\it\sigma},s$ into (3.6) and (3.4) to obtain the physical quantities $\boldsymbol{v}$ and $\boldsymbol{B}$ . Hence, the general non-barotropic magnetohydrodynamic problem is reduced from eight equations (2.1), (2.3), (2.4), (2.5) and the additional constraint (2.2) to a problem of eight first order (in the temporal derivative) unconstrained equations. Moreover, the entire set of equations can be derived from the Lagrangian density $\hat{{\mathcal{L}}}$ .

6 Further simplification

6.1 Elimination of variables

Let us now look at the three last three equations of (3.3). Those describe three comoving quantities which can be written in terms of the generalized Clebsch form given in (3.6) as follows:

(6.1) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle {\displaystyle \frac{\partial {\it\chi}}{\partial t}}+(\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s)\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}=0,\\ \displaystyle {\displaystyle \frac{\partial {\it\eta}}{\partial t}}+(\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s)\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\eta}=0,\\ \displaystyle \frac{\partial s}{\partial t}+(\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}\boldsymbol{{\rm\nabla}}{\it\chi}+{\it\beta}\boldsymbol{{\rm\nabla}}{\it\eta}+{\it\sigma}\boldsymbol{{\rm\nabla}}s)\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}s=0.\end{array}\right\} & \displaystyle\end{eqnarray}$$

Those are algebraic equations for ${\it\alpha},{\it\beta},{\it\sigma}$ which can be solved such that ${\it\alpha},{\it\beta},{\it\sigma}$ can be written as functionals of ${\it\chi},{\it\eta},{\it\nu},s$ , resulting eventually in the description of non-barotropic magnetohydrodynamics in terms of five functions: ${\it\nu},{\it\rho},{\it\chi},{\it\eta},s$ . Let us introduce the notation:

(6.2a−c ) $$\begin{eqnarray}{\it\alpha}_{i}\equiv ({\it\alpha},{\it\beta},{\it\sigma}),\quad {\it\chi}_{i}\equiv ({\it\chi},{\it\eta},s),\quad k_{i}\equiv -\frac{\partial {\it\chi}_{i}}{\partial t}-\boldsymbol{{\rm\nabla}}{\it\nu}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{i},\quad i\in (1,2,3).\end{eqnarray}$$

In terms of the above notation (6.1) takes the form:

(6.3) $$\begin{eqnarray}k_{i}={\it\alpha}_{j}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j},\quad j\in (1,2,3)\end{eqnarray}$$

in which the Einstein summation convention is assumed. Let us define the matrix:

(6.4) $$\begin{eqnarray}\unicode[STIX]{x1D608}_{ij}\equiv \boldsymbol{{\rm\nabla}}{\it\chi}_{i}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\end{eqnarray}$$

obviously this matrix is symmetric since $\unicode[STIX]{x1D608}_{ij}=\unicode[STIX]{x1D608}_{ji}$ . Hence (6.3) takes the form:

(6.5) $$\begin{eqnarray}k_{i}=\unicode[STIX]{x1D608}_{ij}{\it\alpha}_{j},\quad j\in (1,2,3).\end{eqnarray}$$

Provided that the matrix $\unicode[STIX]{x1D608}_{ij}$ is not singular it has an inverse $\unicode[STIX]{x1D608}_{ij}^{-1}$ which can be written as:

(6.6) $$\begin{eqnarray}\unicode[STIX]{x1D608}_{ij}^{-1}=|\unicode[STIX]{x1D63C}|^{-1}\left(\begin{array}{@{}ccc@{}}\unicode[STIX]{x1D608}_{22}\unicode[STIX]{x1D608}_{33}-\unicode[STIX]{x1D608}_{23}^{2} & \unicode[STIX]{x1D608}_{13}\unicode[STIX]{x1D608}_{23}-\unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{33} & \unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{23}-\unicode[STIX]{x1D608}_{13}\unicode[STIX]{x1D608}_{22}\\ \unicode[STIX]{x1D608}_{13}\unicode[STIX]{x1D608}_{23}-\unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{33} & \unicode[STIX]{x1D608}_{11}\unicode[STIX]{x1D608}_{33}-\unicode[STIX]{x1D608}_{13}^{2} & \unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{13}-\unicode[STIX]{x1D608}_{11}\unicode[STIX]{x1D608}_{23}\\ \unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{23}-\unicode[STIX]{x1D608}_{13}\unicode[STIX]{x1D608}_{22} & \unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{13}-\unicode[STIX]{x1D608}_{11}\unicode[STIX]{x1D608}_{23} & \unicode[STIX]{x1D608}_{11}\unicode[STIX]{x1D608}_{22}-\unicode[STIX]{x1D608}_{12}^{2}\end{array}\right).\end{eqnarray}$$

In which the determinant $|\unicode[STIX]{x1D63C}|$ is given by the following equation:

(6.7) $$\begin{eqnarray}|\unicode[STIX]{x1D63C}|=\unicode[STIX]{x1D608}_{11}\unicode[STIX]{x1D608}_{22}\unicode[STIX]{x1D608}_{33}-\unicode[STIX]{x1D608}_{11}\unicode[STIX]{x1D608}_{23}^{2}-\unicode[STIX]{x1D608}_{22}\unicode[STIX]{x1D608}_{13}^{2}-\unicode[STIX]{x1D608}_{33}\unicode[STIX]{x1D608}_{12}^{2}+2\unicode[STIX]{x1D608}_{12}\unicode[STIX]{x1D608}_{13}\unicode[STIX]{x1D608}_{23}.\end{eqnarray}$$

In terms of the above equations the ${\it\alpha}_{i}$ ’s can be calculated as functionals of ${\it\chi}_{i},{\it\nu}$ as follows:

(6.8) $$\begin{eqnarray}{\it\alpha}_{i}[{\it\chi}_{i},{\it\nu}]=\unicode[STIX]{x1D608}_{ij}^{-1}k_{j}.\end{eqnarray}$$

The velocity equation (3.6) can now be written as:

(6.9) $$\begin{eqnarray}\boldsymbol{v}=\boldsymbol{{\rm\nabla}}{\it\nu}+{\it\alpha}_{i}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}=\boldsymbol{{\rm\nabla}}{\it\nu}+\unicode[STIX]{x1D608}_{ij}^{-1}k_{j}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}=\boldsymbol{{\rm\nabla}}{\it\nu}-\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\left(\frac{\partial {\it\chi}_{j}}{\partial t}+\boldsymbol{{\rm\nabla}}{\it\nu}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\right).\end{eqnarray}$$

Provided that the ${\it\chi}_{i}$ is a coordinate basis in three dimensions, we may write:

(6.10) $$\begin{eqnarray}\boldsymbol{{\rm\nabla}}{\it\nu}=\boldsymbol{{\rm\nabla}}{\it\chi}_{n}\frac{\partial {\it\nu}}{\partial {\it\chi}_{n}},\quad n\in (1,2,3).\end{eqnarray}$$

Inserting (6.10) into (6.9) we obtain:

(6.11) $$\begin{eqnarray}\displaystyle \boldsymbol{v} & = & \displaystyle -\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\chi}_{j}}{\partial t}+\boldsymbol{{\rm\nabla}}{\it\nu}-\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\nu}}{\partial {\it\chi}_{n}}\boldsymbol{{\rm\nabla}}{\it\chi}_{n}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\chi}_{j}}{\partial t}+\boldsymbol{{\rm\nabla}}{\it\nu}-\unicode[STIX]{x1D608}_{ij}^{-1}\unicode[STIX]{x1D608}_{jn}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\nu}}{\partial {\it\chi}_{n}}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\chi}_{j}}{\partial t}+\boldsymbol{{\rm\nabla}}{\it\nu}-{\it\delta}_{in}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\nu}}{\partial {\it\chi}_{n}}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\chi}_{j}}{\partial t}+\boldsymbol{{\rm\nabla}}{\it\nu}-\boldsymbol{{\rm\nabla}}{\it\chi}_{n}\frac{\partial {\it\nu}}{\partial {\it\chi}_{n}}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\chi}_{j}}{\partial t}\end{eqnarray}$$

in the above ${\it\delta}_{in}$ is a Kronecker delta. Thus the velocity $\boldsymbol{v}[{\it\chi}_{i}]$ is a functional of ${\it\chi}_{i}$ only and is independent of  ${\it\nu}$ .

6.2 Lagrangian density and variational analysis

Let us now rewrite the Lagrangian density $\hat{{\mathcal{L}}}[{\it\chi}_{i},{\it\nu},{\it\rho}]$ given in (5.3) in terms of the new variables:

(6.12) $$\begin{eqnarray}\hat{{\mathcal{L}}}[{\it\chi}_{i},{\it\nu},{\it\rho}]=-{\it\rho}\left[\frac{\partial {\it\nu}}{\partial t}+{\it\alpha}_{k}[{\it\chi}_{i},{\it\nu}]\frac{\partial {\it\chi}_{k}}{\partial t}+{\it\varepsilon}({\it\rho},{\it\chi}_{3})+\frac{1}{2}\boldsymbol{v}[{\it\chi}_{i}]^{2}\right]-\frac{1}{8{\rm\pi}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2})^{2}.\end{eqnarray}$$

Let us calculate the variational derivative of $\hat{{\mathcal{L}}}[{\it\chi}_{i},{\it\nu},{\it\rho}]$ with respect to ${\it\chi}_{i}$ , this will result in:

(6.13) $$\begin{eqnarray}{\it\delta}_{{\it\chi}_{i}}\hat{{\mathcal{L}}}=-{\it\rho}\left[{\it\delta}_{{\it\chi}_{i}}{\it\alpha}_{k}\frac{\partial {\it\chi}_{k}}{\partial t}+{\it\alpha}_{\underline{i}}\frac{\partial {\it\delta}{\it\chi}_{\underline{i}}}{\partial t}+{\it\delta}_{{\it\chi}_{i}}{\it\varepsilon}({\it\rho},{\it\chi}_{3})+{\it\delta}_{{\it\chi}_{i}}\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{v}\right]-\frac{\boldsymbol{B}}{4{\rm\pi}}\boldsymbol{\cdot }{\it\delta}_{{\it\chi}_{i}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2}),\end{eqnarray}$$

in which the summation convention is not applied if the index is underlined. However, due to (6.9) we may write:

(6.14) $$\begin{eqnarray}{\it\delta}_{{\it\chi}_{i}}\boldsymbol{v}={\it\delta}_{{\it\chi}_{i}}{\it\alpha}_{k}\boldsymbol{{\rm\nabla}}{\it\chi}_{k}+{\it\alpha}_{\underline{i}}\boldsymbol{{\rm\nabla}}{\it\delta}{\it\chi}_{\underline{i}}.\end{eqnarray}$$

Inserting (6.14) into (6.13) and rearranging the terms we obtain:

(6.15) $$\begin{eqnarray}\displaystyle {\it\delta}_{{\it\chi}_{i}}\hat{{\mathcal{L}}} & = & \displaystyle -{\it\rho}\left[{\it\delta}_{{\it\chi}_{i}}{\it\alpha}_{k}\left(\frac{\partial {\it\chi}_{k}}{\partial t}+\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{k}\right)+{\it\alpha}_{\underline{i}}\left(\frac{\partial {\it\delta}{\it\chi}_{\underline{i}}}{\partial t}+\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\delta}{\it\chi}_{\underline{i}}\right)+{\it\delta}_{{\it\chi}_{i}}{\it\varepsilon}({\it\rho},{\it\chi}_{3})\right]\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\boldsymbol{B}}{4{\rm\pi}}\boldsymbol{\cdot }{\it\delta}_{{\it\chi}_{i}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2}).\end{eqnarray}$$

Now by construction $\boldsymbol{v}$ satisfies (6.1) and hence $\partial {\it\chi}_{k}/\partial t+\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{k}=0$ , this leads to:

(6.16) $$\begin{eqnarray}{\it\delta}_{{\it\chi}_{i}}\hat{{\mathcal{L}}}=-{\it\rho}\left[{\it\alpha}_{\underline{i}}\frac{\text{d}{\it\delta}{\it\chi}_{\underline{i}}}{\text{d}t}+{\it\delta}_{{\it\chi}_{i}}{\it\varepsilon}({\it\rho},{\it\chi}_{3})\right]-\frac{\boldsymbol{B}}{4{\rm\pi}}\boldsymbol{\cdot }{\it\delta}_{{\it\chi}_{i}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2}).\end{eqnarray}$$

From now on the derivation proceeds as in (3.10), (3.11), (3.13) resulting in (3.12), (3.14) and will not be repeated. The difference is that now ${\it\alpha},{\it\beta}$ and ${\it\sigma}$ are not independent quantities, rather they depend through (6.8) on the derivatives of ${\it\chi}_{i},{\it\nu}$ . Thus, (3.10), (3.11), (3.13) are not first-order equations in time but are second-order equations. Now let us calculate the variational derivative with respect to ${\it\nu}$ this will result in the expression:

(6.17) $$\begin{eqnarray}{\it\delta}_{{\it\nu}}\hat{{\mathcal{L}}}=-{\it\rho}\left[\frac{\partial {\it\delta}{\it\nu}}{\partial t}+{\it\delta}_{{\it\nu}}{\it\alpha}_{n}\frac{\partial {\it\chi}_{n}}{\partial t}\right].\end{eqnarray}$$

However, ${\it\delta}_{{\it\nu}}{\it\alpha}_{k}$ can be calculated from (6.8):

(6.18) $$\begin{eqnarray}{\it\delta}_{{\it\nu}}{\it\alpha}_{n}=\unicode[STIX]{x1D608}_{nj}^{-1}{\it\delta}_{{\it\nu}}k_{j}=-\unicode[STIX]{x1D608}_{nj}^{-1}\boldsymbol{{\rm\nabla}}{\it\delta}{\it\nu}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}.\end{eqnarray}$$

Inserting the above equation into (6.17):

(6.19) $$\begin{eqnarray}{\it\delta}_{{\it\nu}}\hat{{\mathcal{L}}}=-{\it\rho}\left[\frac{\partial {\it\delta}{\it\nu}}{\partial t}-\unicode[STIX]{x1D608}_{nj}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\frac{\partial {\it\chi}_{n}}{\partial t}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\delta}{\it\nu}\right]=-{\it\rho}\left[\frac{\partial {\it\delta}{\it\nu}}{\partial t}+\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\delta}{\it\nu}\right]=-{\it\rho}\frac{\text{d}{\it\delta}{\it\nu}}{\text{d}t}.\end{eqnarray}$$

The above equation can be put in the form:

(6.20) $$\begin{eqnarray}{\it\delta}_{{\it\nu}}\hat{{\mathcal{L}}}={\it\delta}{\it\nu}\left[\frac{\partial {\it\rho}}{\partial t}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}\boldsymbol{v})\right]-\frac{\partial ({\it\rho}{\it\delta}{\it\nu})}{\partial t}-\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}\boldsymbol{v}{\it\delta}{\it\nu}).\end{eqnarray}$$

This obviously leads to the continuity equation (2.3) and some boundary terms in space and time. The variational derivative with respect to ${\it\rho}$ is trivial and the analysis is identical to the one in (3.7) leading to (3.8). To conclude this subsection let us summarize the equations of non-barotropic magnetohydrodynamics:

(6.21) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\begin{array}{@{}c@{}}\displaystyle {\displaystyle \frac{\text{d}{\it\nu}}{\text{d}t}}={\displaystyle \frac{1}{2}}\boldsymbol{v}^{2}-w,\\ \displaystyle {\displaystyle \frac{\partial {\it\rho}}{\partial t}}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\rho}\boldsymbol{v})=0,\\ \displaystyle {\displaystyle \frac{\text{d}{\it\sigma}}{\text{d}t}}=T,\\ \displaystyle \frac{\text{d}{\it\alpha}}{\text{d}t}={\displaystyle \frac{\boldsymbol{{\rm\nabla}}{\it\eta}\boldsymbol{\cdot }\boldsymbol{J}}{{\it\rho}}},\\ \displaystyle {\displaystyle \frac{\text{d}{\it\beta}}{\text{d}t}}=-{\displaystyle \frac{\boldsymbol{{\rm\nabla}}{\it\chi}\boldsymbol{\cdot }\boldsymbol{J}}{{\it\rho}}},\end{array}\right\} & \displaystyle\end{eqnarray}$$

in which ${\it\alpha},{\it\beta},{\it\sigma},\boldsymbol{v}$ are functionals of ${\it\chi},{\it\eta},s,{\it\nu}$ as described above. It is easy to show as in (4.3) that those variational equations are equivalent to the physical equations.

6.3 Lagrangian and Hamiltonian densities

Let us put the Lagrangian density of equation (6.12) in a slightly more explicit form. First us look at the term $\boldsymbol{v}^{2}$ :

(6.22) $$\begin{eqnarray}\boldsymbol{v}^{2}=\unicode[STIX]{x1D608}_{ij}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{i}\frac{\partial {\it\chi}_{j}}{\partial t}\unicode[STIX]{x1D608}_{mn}^{-1}\boldsymbol{{\rm\nabla}}{\it\chi}_{m}\frac{\partial {\it\chi}_{n}}{\partial t}=\unicode[STIX]{x1D608}_{ij}^{-1}\unicode[STIX]{x1D608}_{mn}^{-1}\unicode[STIX]{x1D608}_{im}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{n}}{\partial t}=\unicode[STIX]{x1D608}_{jn}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{n}}{\partial t}\end{eqnarray}$$

in the above we use (6.11) and (6.4). Next let us look at the expression:

(6.23) $$\begin{eqnarray}\displaystyle {\it\alpha}_{k}[{\it\chi}_{i},{\it\nu}]\frac{\partial {\it\chi}_{k}}{\partial t} & = & \displaystyle \unicode[STIX]{x1D608}_{kj}^{-1}k_{j}\frac{\partial {\it\chi}_{k}}{\partial t}=-\left(\frac{\partial {\it\chi}_{j}}{\partial t}+\boldsymbol{{\rm\nabla}}{\it\nu}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\right)\unicode[STIX]{x1D608}_{kj}^{-1}\frac{\partial {\it\chi}_{k}}{\partial t}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{jk}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{k}}{\partial t}-\boldsymbol{{\rm\nabla}}{\it\nu}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\unicode[STIX]{x1D608}_{kj}^{-1}\frac{\partial {\it\chi}_{k}}{\partial t}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{jk}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{k}}{\partial t}-\frac{\partial {\it\nu}}{\partial {\it\chi}_{m}}\boldsymbol{{\rm\nabla}}{\it\chi}_{m}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\chi}_{j}\unicode[STIX]{x1D608}_{kj}^{-1}\frac{\partial {\it\chi}_{k}}{\partial t}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{jk}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{k}}{\partial t}-\frac{\partial {\it\nu}}{\partial {\it\chi}_{m}}\unicode[STIX]{x1D608}_{mj}\unicode[STIX]{x1D608}_{kj}^{-1}\frac{\partial {\it\chi}_{k}}{\partial t}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{jk}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{k}}{\partial t}-\frac{\partial {\it\nu}}{\partial {\it\chi}_{m}}{\it\delta}_{km}\frac{\partial {\it\chi}_{k}}{\partial t}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D608}_{jk}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{k}}{\partial t}-\frac{\partial {\it\nu}}{\partial {\it\chi}_{m}}\frac{\partial {\it\chi}_{m}}{\partial t}.\end{eqnarray}$$

Inserting (6.22) and (6.23) into (6.12) leads to a Lagrangian density of a more standard quadratic form:

(6.24) $$\begin{eqnarray}\displaystyle \hat{{\mathcal{L}}}[{\it\chi}_{i},{\it\nu},{\it\rho}]={\it\rho}\left[\frac{1}{2}\unicode[STIX]{x1D608}_{jn}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{n}}{\partial t}+\frac{\partial {\it\nu}}{\partial {\it\chi}_{m}}\frac{\partial {\it\chi}_{m}}{\partial t}-\frac{\partial {\it\nu}}{\partial t}-{\it\varepsilon}({\it\rho},{\it\chi}_{3})\right]-\frac{1}{8{\rm\pi}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2})^{2}. & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

In which $\unicode[STIX]{x1D608}_{jn}^{-1}$ plays the rule of a ‘metric’. The Lagrangian is thus composed of a kinetic terms which is quadratic in the temporal derivatives, a ‘gyroscopic’ term which is linear in the temporal derivative and a potential term which is independent of the temporal derivative.

In order to obtain a Hamiltonian density of a more convenient form we will add a temporal derivative to the Lagrangian density $\hat{{\mathcal{L}}}$ , this will not change the dynamical equations as is well known, hence we can write:

(6.25) $$\begin{eqnarray}\displaystyle \tilde{{\mathcal{L}}}[{\it\chi}_{i},{\it\nu},{\it\rho}] & = & \displaystyle \hat{{\mathcal{L}}}+\frac{\partial ({\it\rho}{\it\nu})}{\partial t}={\it\nu}\frac{\partial {\it\rho}}{\partial t}+{\it\rho}\left[\frac{1}{2}\unicode[STIX]{x1D608}_{jn}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{n}}{\partial t}+\frac{\partial {\it\nu}}{\partial {\it\chi}_{m}}\frac{\partial {\it\chi}_{m}}{\partial t}-{\it\varepsilon}({\it\rho},{\it\chi}_{3})\right]\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{8{\rm\pi}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2})^{2}.\end{eqnarray}$$

From this we obtain the following canonical momenta:

(6.26) $$\begin{eqnarray}p_{{\it\rho}}=\frac{\partial \tilde{{\mathcal{L}}}}{\partial \dot{{\it\rho}}}={\it\nu},\quad \dot{{\it\rho}}\equiv \frac{\partial {\it\rho}}{\partial t}.\end{eqnarray}$$

And:

(6.27) $$\begin{eqnarray}p_{{\it\chi}_{k}}=\frac{\partial \tilde{{\mathcal{L}}}}{\partial \dot{{\it\chi}}_{k}}={\it\rho}\left[\unicode[STIX]{x1D608}_{kj}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}+\frac{\partial {\it\nu}}{\partial {\it\chi}_{k}}\right]=-{\it\rho}{\it\alpha}_{k},\quad \dot{{\it\chi}}_{k}\equiv \frac{\partial {\it\chi}_{k}}{\partial t}.\end{eqnarray}$$

The Hamiltonian density can be now calculated as follows:

(6.28) $$\begin{eqnarray}\displaystyle {\mathcal{H}}[{\it\chi}_{i},{\it\rho},p_{{\it\chi}_{k}},p_{{\it\rho}}] & = & \displaystyle p_{{\it\chi}_{k}}\dot{{\it\chi}}_{k}+p_{{\it\rho}}\dot{{\it\rho}}-\tilde{{\mathcal{L}}}\nonumber\\ \displaystyle & = & \displaystyle {\it\rho}\left[\frac{1}{2}\unicode[STIX]{x1D608}_{jn}^{-1}\frac{\partial {\it\chi}_{j}}{\partial t}\frac{\partial {\it\chi}_{n}}{\partial t}+{\it\varepsilon}({\it\rho},{\it\chi}_{3})\right]+\frac{1}{8{\rm\pi}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2})^{2}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}{\it\rho}^{-1}\unicode[STIX]{x1D608}_{kl}p_{{\it\chi}_{k}}p_{{\it\chi}_{l}}-p_{{\it\chi}_{k}}\boldsymbol{{\rm\nabla}}{\it\chi}_{k}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}p_{{\it\rho}}+\frac{1}{2}{\it\rho}(\boldsymbol{{\rm\nabla}}p_{{\it\rho}})^{2}\nonumber\\ \displaystyle & & \displaystyle +\,{\it\rho}{\it\varepsilon}({\it\rho},{\it\chi}_{3})+\frac{1}{8{\rm\pi}}(\boldsymbol{{\rm\nabla}}{\it\chi}_{1}\times \boldsymbol{{\rm\nabla}}{\it\chi}_{2})^{2}.\end{eqnarray}$$

Hence by virtue of (6.22) the Hamiltonian density is equal to the energy density of the flow:

(6.29) $$\begin{eqnarray}{\mathcal{H}}=\frac{1}{2}{\it\rho}\boldsymbol{v}^{2}+{\it\rho}{\it\varepsilon}+\frac{\boldsymbol{B}^{2}}{8{\rm\pi}}.\end{eqnarray}$$

The Hamilton equations are now:

(6.30) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{{\it\delta}{\mathcal{H}}}{{\it\delta}p_{{\it\rho}}}=\dot{{\it\rho}} & \displaystyle\end{eqnarray}$$

(6.31) $$\begin{eqnarray}\displaystyle & \displaystyle -\frac{{\it\delta}{\mathcal{H}}}{{\it\delta}{\it\rho}}={\dot{p}}_{{\it\rho}} & \displaystyle\end{eqnarray}$$

(6.32) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{{\it\delta}{\mathcal{H}}}{{\it\delta}p_{{\it\chi}_{k}}}=\dot{{\it\chi}}_{k} & \displaystyle\end{eqnarray}$$

(6.33) $$\begin{eqnarray}\displaystyle & \displaystyle -\frac{{\it\delta}{\mathcal{H}}}{{\it\delta}{\it\chi}_{k}}={\dot{p}}_{{\it\chi}_{k}}. & \displaystyle\end{eqnarray}$$

Equation (6.30) is equivalent to the continuity equation (2.3). Equation (6.31) is equivalent to the Bernoulli-type equation (3.8). Equation (6.32) is equivalent to (6.1) and thus is satisfied identically. Equation (6.32) is equivalent to (3.12) and (3.14). Hence the above equations are the same as (6.21) but in Hamiltonian presentation.