3.1. Euler's equations

We shall now show that a velocity field given by (3.6), such that the equations for $\alpha , \beta , \chi , \eta , \nu , \sigma , s$ satisfy the corresponding equations (3.3), (3.8), (3.11a,b), (3.13) must satisfy Euler's equations. Let us calculate the material derivative of $\boldsymbol {v}$ as follows:

(3.14)\begin{align} \frac{\textrm{d}\boldsymbol{v}}{\textrm{d} t} &= \frac{\textrm{d}\boldsymbol{\nabla} \nu}{\textrm{d} t} + \frac{\textrm{d}\alpha}{\textrm{d} t} \boldsymbol{\nabla} \chi + \alpha \frac{\textrm{d}\boldsymbol{\nabla} \chi}{\textrm{d} t} + \frac{\textrm{d}\beta}{\textrm{d} t} \boldsymbol{\nabla} \eta + \beta \frac{\textrm{d}\boldsymbol{\nabla} \eta}{\textrm{d} t} +\frac{\textrm{d}\sigma}{\textrm{d} t} \boldsymbol{\nabla} s + \sigma \frac{\textrm{d}\boldsymbol{\nabla} s}{\textrm{d} t}. \end{align}

It can be easily shown that

(3.15)\begin{equation} \left.\begin{gathered} \frac{\textrm{d}\boldsymbol{\nabla} \nu}{\textrm{d} t} = \boldsymbol{\nabla} \frac{\textrm{d} \nu}{\textrm{d} t}- \boldsymbol{\nabla} v_k \frac{\partial \nu}{\partial x_k} = \boldsymbol{\nabla} \left(\frac{1}{2} \boldsymbol{v}^2 - w\right)- \boldsymbol{\nabla} v_k \frac{\partial \nu}{\partial x_k}, \\ \frac{\textrm{d}\boldsymbol{\nabla} \eta}{\textrm{d} t} = \boldsymbol{\nabla} \frac{\textrm{d} \eta}{\textrm{d} t}- \boldsymbol{\nabla} v_k \frac{\partial \eta}{\partial x_k} ={-} \boldsymbol{\nabla} v_k \frac{\partial \eta}{\partial x_k}, \\ \frac{\textrm{d}\boldsymbol{\nabla} \chi}{\textrm{d} t} = \boldsymbol{\nabla} \frac{\textrm{d} \chi}{\textrm{d} t}- \boldsymbol{\nabla} v_k \frac{\partial \chi}{\partial x_k} ={-} \boldsymbol{\nabla} v_k \frac{\partial \chi}{\partial x_k}, \\ \frac{\textrm{d}\boldsymbol{\nabla} s}{\textrm{d} t} = \boldsymbol{\nabla} \frac{\textrm{d} s}{\textrm{d} t}- \boldsymbol{\nabla} v_k \frac{\partial s}{\partial x_k} ={-} \boldsymbol{\nabla} v_k \frac{\partial s}{\partial x_k}. \end{gathered}\right\}\end{equation}

In which $x_k$ is a Cartesian coordinate and a summation convention is assumed. Inserting the results from (3.15) and (3.3) into (3.14) yields

(3.16)\begin{align} \frac{\textrm{d}\boldsymbol{v}}{\textrm{d} t} &={-} \boldsymbol{\nabla} v_k \left(\frac{\partial \nu}{\partial x_k} + \alpha \frac{\partial \chi}{\partial x_k} + \beta \frac{\partial \eta}{\partial x_k} + \sigma \frac{\partial s}{\partial x_k}\right) +\boldsymbol{\nabla} \left(\frac{1}{2} \boldsymbol{v}^2 - w\right)+ T \boldsymbol{\nabla} s \nonumber\\ &\quad +\frac{1}{\rho} ((\boldsymbol{\nabla} \eta \boldsymbol{\cdot} \boldsymbol{J})\boldsymbol{\nabla} \chi -(\boldsymbol{\nabla} \chi \boldsymbol{\cdot} \boldsymbol{J})\boldsymbol{\nabla} \eta) \nonumber\\ &={-} \boldsymbol{\nabla} v_k v_k + \boldsymbol{\nabla} \left(\frac{1}{2} \boldsymbol{v}^2 - w\right) + T \boldsymbol{\nabla} s +\frac{1}{\rho} \boldsymbol{J} \times (\boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \eta) \nonumber\\ &={-} \frac{\boldsymbol{\nabla} p}{\rho} + \frac{1}{\rho} \boldsymbol{J} \times \boldsymbol{B}. \end{align}

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 MHD can be derived from the above action without restricting the variations in any way except on the relevant boundaries and cuts.

3.2. Local non-barotropic cross-helicity

The function $\nu$, whose material derivative is given in (3.8), can be multiple valued because only its gradient appears in the velocity (3.6). However, the discontinuity, $[\nu ]$, of $\nu$ is conserved,

(3.17)\begin{equation} \displaystyle\frac{\textrm{d}{[\nu]}}{\textrm{d} t} =0, \end{equation}

since the terms on the right-hand side of (3.8) describe physical quantities and hence are single valued. A similar equation also holds for barotropic fluid dynamics and barotropic MHD (Yahalom & Lynden-Bell Reference Yahalom and Lynden-Bell2008; Yahalom Reference Yahalom2010, Reference Yahalom2013a).

We now substitute the expressions for ${\boldsymbol {B}}$ and ${\boldsymbol {v}}$ given by (3.4) and (3.6), respectively, into the formula $H_{C} \equiv \int \boldsymbol {B}\cdot \boldsymbol {v}\,{\textrm {d}}^{3} x$ for the cross-helicity (see (1.1)) to obtain

(3.18)\begin{equation} H_{C} = \int {\textrm{d}} \varPhi [\nu] + \int {\textrm{d}} \varPhi \oint \sigma \,\textrm{d} s, \end{equation}

where the closed line integral is taken along a magnetic field line. Furthermore, ${\textrm {d}}\varPhi =\boldsymbol {B}\cdot \textrm {d} \boldsymbol {S}=({\boldsymbol {\nabla }}\chi \times {\boldsymbol {\nabla }}\eta ) \boldsymbol {\cdot } \textrm {d} \boldsymbol {S}={\textrm {d}}\chi \,{\textrm {d}}\eta$ is a magnetic flux element which is comoving as governed by (2.1) and ${\textrm {d}}\boldsymbol {S}$ is an infinitesimal area element. Although the cross-helicity is not conserved for non-barotropic flows, inspection of the right-hand side of (3.18) reveals that it is made of a sum of two terms. One term is conserved, as both ${\textrm {d}}\varPhi$ and $[\nu ]$ are comoving, and the other is not. This suggests the following definition for the non-barotropic cross-helicity:

(3.19)\begin{equation} H_{CNB} \equiv \int {\textrm{d}} \varPhi [\nu] = H_{C} - \int {\textrm{d}} \varPhi \oint \sigma\,{\textrm{d}} s. \end{equation}

It can be written in the more conventional form,

(3.20)\begin{equation} H_{CNB} = \int \boldsymbol{B} \cdot \boldsymbol{v}_t \,{\textrm{d}}^{3} x, \end{equation}

in which the topological velocity field is defined as

(3.21)\begin{equation} \boldsymbol{v}_t = \boldsymbol{v} - \sigma \boldsymbol{\nabla} s. \end{equation}

It should be noted that $H_{CNB}$ is conserved even for an MHD not satisfying the Sakurai topological constraint given in (3.4), provided that we have a field $\sigma$ satisfying the equation ${\textrm {d}{\sigma }}/{\textrm {d} t} = T$. This can be verified by direct derivation using only the equation of motion and the sigma equation. Thus the non-barotropic cross-helicity conservation law,

(3.22)\begin{equation} \displaystyle\frac{{{\textrm{d}}}{H_{CNB}}}{{{\textrm{d}}}{t}} = 0, \end{equation}

is more general than the variational principle described by (3.50) as follows from a direct computation using (2.1) and (2.3)–(2.5). Also note that, for a constant specific entropy $s$, we obtain $H_{CNB}=H_{C}$ and the non-barotropic cross-helicity reduces to the standard barotropic cross-helicity. The local form of (3.22) describing the evolution of $H_{CNB}$ per unit volume was described by (Webb et al. Reference Webb, Dasgupta, McKenzie, Hu and Zank2014a,Reference Webb, Dasgupta, McKenzie, Hu and Zankb). To conclude, we introduce also a local topological conservation law in the spirit of Yahalom (Yahalom Reference Yahalom2013a) which is the non-barotropic cross-helicity per unit of magnetic flux. This quantity which is equal to the discontinuity, $[\nu ]$, of $\nu$ is conserved and can be written as a sum of the barotropic cross-helicity per unit flux and the closed line integral of $s {\textrm {d}} \sigma$ along a magnetic field line, namely

(3.23)\begin{equation} [\nu]= \displaystyle\frac{{{\textrm{d}}}{H_{CNB}}}{{{\textrm{d}}}{\varPhi}} = \displaystyle\frac{{{\textrm{d}}}{H_{C}}}{{{\textrm{d}}}{\varPhi}} + \oint s \,{\textrm{d}} \sigma. \end{equation}

3.3. Simplified action

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

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

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

(3.25)\begin{equation} \left.\begin{aligned} &\hspace{3pc}{\mathcal{L}} = \hat{\mathcal{L}} + {\mathcal{L}}_{\boldsymbol{v}}+ {\mathcal{L}}_{\boldsymbol{B}}+{\mathcal{L}}_{boundary}, \\ \hat{\mathcal{L}} & \equiv{-}\rho \left[\frac{\partial{\nu}}{\partial t} +\alpha \frac{\partial{\chi}}{\partial t} +\beta \frac{\partial{\eta}}{\partial t}+ \sigma \frac{\partial{s}}{\partial t}+\varepsilon (\rho,s) \right.\\ & \quad \left.+\frac{1}{2} (\boldsymbol{\nabla} \nu + \alpha \boldsymbol{\nabla} \chi + \beta \boldsymbol{\nabla} \eta + \sigma \boldsymbol{\nabla} s )^2 \right] \\ & \quad -\frac{1}{8 {\rm \pi}}(\boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \eta)^2 ,\\ &\hspace{4pc}{\mathcal{L}}_{\boldsymbol{v}} \equiv \frac{1}{2}\rho (\boldsymbol{v}-\hat{\boldsymbol{v}})^2, \\ &\hspace{4pc}{\mathcal{L}}_{\boldsymbol{B}} \equiv \frac{1}{8 {\rm \pi}} (\boldsymbol{B}-\hat{\boldsymbol{B}})^2, \\ &\hspace{3pc}{\mathcal{L}}_{boundary} \equiv \frac{\partial{(\nu \rho)}}{\partial t} + \boldsymbol{\nabla} \boldsymbol{\cdot} (\nu \rho \boldsymbol{v} ). \end{aligned}\right\} \end{equation}

The only term containing $\boldsymbol {v}$ is ${\mathcal {L}}_{\boldsymbol {v}}$ (${\mathcal {L}}_{boundary}$ also depends on $\boldsymbol {v}$ but being a boundary term in space and time it does not contribute to the derived equations), 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 notice that the term ${\mathcal {L}}_{boundary}$ contains only complete partial derivatives and thus can not contribute to the equations, although it can change the boundary conditions. Hence we see that (3.3), (3.8), (3.11a,b) and (3.13) can be derived using the Lagrangian density,

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

in which $\hat {\boldsymbol {v}}$ replaces $\boldsymbol {v}$ and $\hat {\boldsymbol {B}}$ replaces $\boldsymbol {B}$ in the relevant equations. Furthermore, after integrating the eight equations (3.3), (3.8), (3.11a,b), (3.13) we can insert the potentials $\alpha ,\beta ,\chi ,\eta ,\nu ,\sigma ,s$ 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 $\hat {\mathcal {L}}$.

3.4. Further simplification

3.4.1. Elimination of variables

Let us now look at the last three equations of (3.3) (Yahalom Reference Yahalom2016a,Reference Yahalomb). Those describe three comoving quantities which can be written in terms of the generalized Clebsch form given in (3.6) as follows:

(3.27)\begin{equation} \left.\begin{gathered} \dfrac{\partial \chi}{\partial t} + (\boldsymbol{\nabla} \nu + \alpha \boldsymbol{\nabla} \chi + \beta \boldsymbol{\nabla} \eta + \sigma \boldsymbol{\nabla} s) \boldsymbol{\cdot} \boldsymbol{\nabla} \chi = 0 \\ \dfrac{\partial \eta}{\partial t} + (\boldsymbol{\nabla} \nu + \alpha \boldsymbol{\nabla} \chi + \beta \boldsymbol{\nabla} \eta + \sigma \boldsymbol{\nabla} s) \boldsymbol{\cdot} \boldsymbol{\nabla} \eta = 0 \\ \dfrac{\partial s}{\partial t} + (\boldsymbol{\nabla} \nu + \alpha \boldsymbol{\nabla} \chi + \beta \boldsymbol{\nabla} \eta + \sigma \boldsymbol{\nabla} s) \boldsymbol{\cdot} \boldsymbol{\nabla} s = 0 \end{gathered}\right\}. \end{equation}

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

(3.28)\begin{equation} \alpha_i \equiv (\alpha, \beta, \sigma), \chi_i\equiv (\chi,\eta,s), k_i \equiv{-}\frac{\partial \chi_i}{\partial t} - \boldsymbol{\nabla} \nu \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_i ,\end{equation}

$i\in (1,2,3)$. In terms of the above notation (3.27) takes the form

(3.29)\begin{equation} k_i =\alpha_j \boldsymbol{\nabla} \chi_i \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_j,\quad j\in(1,2,3) \end{equation}

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

(3.30)\begin{equation} A_{ij} \equiv \boldsymbol{\nabla} \chi_i \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_j ,\end{equation}

where obviously this matrix is symmetric since $A_{ij}=A_{ji}$. Hence (3.29) takes the form

(3.31)\begin{equation} k_i = A_{ij} \alpha_j,\quad j\in(1,2,3). \end{equation}

Provided that the matrix $A_{ij}$ is not singular it has an inverse $A^{-1}_{ij}$ which can be written as

(3.32)\begin{align} A^{{-}1}_{ij} &= \left|A\right|^{{-}1} \left(\begin{array}{@{}ccc@{}} A_{22} A_{33}-A_{23}^2 & A_{13} A_{23}-A_{12} A_{33} & A_{12} A_{23}-A_{13} A_{22} \\ A_{13} A_{23}-A_{12} A_{33} & A_{11} A_{33}-A_{13}^2 & A_{12} A_{13}-A_{11} A_{23} \\ A_{12} A_{23}-A_{13} A_{22} & A_{12} A_{13}-A_{11} A_{23} & A_{11} A_{22}-A_{12}^2 \end{array}\right), \end{align}

in which the determinant $|A|$ is given by the following equation:

(3.33)\begin{align} \left|A\right| &= A_{11} A_{22} A_{33}-A_{11} A_{23}^2-A_{22} A_{13}^2 -A_{33} A_{12}^2 +2 A_{12} A_{13} A_{23} . \end{align}

In terms of the above equations the $\alpha _i$ can be calculated as functionals of $\chi _i,\nu$ as follows:

(3.34)\begin{equation} \alpha_i [\chi_i,\nu]= A^{{-}1}_{ij} k_j. \end{equation}

The velocity equation (3.6) can now be written as

(3.35)\begin{align} \boldsymbol{v} &= \boldsymbol{\nabla} \nu + \alpha_i \boldsymbol{\nabla} \chi_i= \boldsymbol{\nabla} \nu + A^{{-}1}_{ij} k_j \boldsymbol{\nabla} \chi_i \nonumber\\ &= \boldsymbol{\nabla} \nu - A^{{-}1}_{ij}\boldsymbol{\nabla} \chi_i \left(\frac{\partial \chi_j}{\partial t} + \boldsymbol{\nabla} \nu \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_j\right). \end{align}

Provided that the $\chi _i$ is a coordinate basis in three dimensions, we may write

(3.36)\begin{equation} \boldsymbol{\nabla} \nu= \boldsymbol{\nabla} \chi_n \frac{\partial \nu}{\partial \chi_n},\quad n\in(1,2,3). \end{equation}

Inserting (3.36) into (3.35) we obtain

(3.37)\begin{equation} \boldsymbol{v} ={-} A^{{-}1}_{ij}\boldsymbol{\nabla} \chi_i \frac{\partial \chi_j}{\partial t} \end{equation}

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

3.5. Lagrangian density and variational analysis

Let us now rewrite the Lagrangian density $\hat {\mathcal {L}}[\chi _i,\nu ,\rho ]$ given in (3.26) in terms of the new variables as

(3.38) \begin{align} \hat{\mathcal{L}}[\chi_i,\nu,\rho] &={-}\rho \left[\frac{\partial{\nu}}{\partial t} + \alpha_k [\chi_i,\nu] \frac{\partial{\chi_k}}{\partial t} + \varepsilon (\rho,\chi_3) + \frac{1}{2} \boldsymbol{v} [\chi_i]^2\right] \nonumber\\ &\quad -\frac{1}{8 {\rm \pi}}(\boldsymbol{\nabla} \chi_1 \times \boldsymbol{\nabla} \chi_2)^2. \end{align}

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

(3.39)\begin{align} \delta_{\chi_i}\hat{\mathcal{L}} &={-}\rho \left[\delta_{\chi_i} \alpha_k \frac{\partial{\chi_k}}{\partial t} + \alpha_{\underline{i}} \frac{\partial \delta \chi_{\underline{i}}}{\partial t} +\delta_{\chi_i} \varepsilon (\rho,\chi_3) + \delta_{\chi_i}\boldsymbol{v} \cdot \boldsymbol{v}\right] \nonumber\\ &\quad -\frac{ \boldsymbol{B}} {4 {\rm \pi}} \cdot \delta_{\chi_i} (\boldsymbol{\nabla} \chi_1 \times \boldsymbol{\nabla} \chi_2) \end{align}

in which the summation convention is not applied if the index is underlined. However, due to (3.35) we may write

(3.40)\begin{equation} \delta_{\chi_i}\boldsymbol{v}= \delta_{\chi_i} \alpha_k \boldsymbol{\nabla} \chi_k + \alpha_{\underline{i}} \boldsymbol{\nabla} \delta \chi_{\underline{i}}. \end{equation}

Inserting (3.40) into (3.39) and rearranging the terms we obtain

(3.41)\begin{align} \delta_{\chi_i}\hat{\mathcal{L}} &={-}\rho \left[\delta_{\chi_i} \alpha_k \left(\frac{\partial{\chi_k}}{\partial t} + \boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_k\right) +\alpha_{\underline{i}} \left(\frac{\partial \delta \chi_{\underline{i}}}{\partial t} +\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \delta \chi_{\underline{i}}\right) + \delta_{\chi_i} \varepsilon (\rho,\chi_3)\right] \nonumber\\ &\quad -\frac{ \boldsymbol{B}} {4 {\rm \pi}} \cdot \delta_{\chi_i} (\boldsymbol{\nabla} \chi_1 \times \boldsymbol{\nabla} \chi_2). \end{align}

Now by construction $\boldsymbol {v}$ satisfies (3.27) and hence ${\partial {\chi _k}}/{\partial t} + \boldsymbol {v} \boldsymbol {\cdot } \boldsymbol {\nabla } \chi _k = 0$, this leads to

(3.42)\begin{equation} \delta_{\chi_i}\hat{\mathcal{L}} ={-}\rho \left[\alpha_{\underline{i}} \frac{\textrm{d} \delta \chi_{\underline{i}}}{\textrm{d} t} + \delta_{\chi_i} \varepsilon (\rho,\chi_3) \right] - \frac{ \boldsymbol{B}} {4 {\rm \pi}} \boldsymbol{\cdot} \delta_{\chi_i} (\boldsymbol{\nabla} \chi_1 \times \boldsymbol{\nabla} \chi_2). \end{equation}

From now on the derivation proceeds as in (3.9), (3.10), (3.12) resulting in (3.11a,b), (3.13) and will not be repeated. The difference is that now $\alpha , \beta$ and $\sigma$ are not independent quantities, rather they depend through (3.34) on the derivatives of $\chi _i,\nu$. Thus, (3.9), (3.10), (3.12) are not first-order equations in time but are second-order equations. Now let us calculate the variational derivative with respect to $\nu$, this will result in the expression

(3.43)\begin{equation} \delta_{\nu} \hat{\mathcal{L}} ={-}\rho \left[\dfrac{\partial{\delta \nu}}{\partial t} + \delta_{\nu} \alpha_n \dfrac{\partial{\chi_n}}{\partial t}\right]. \end{equation}

However, $\delta _{\nu } \alpha _k$ can be calculated from (3.34) as follows:

(3.44)\begin{equation} \delta_{\nu} \alpha_n = A^{{-}1}_{nj} \delta_{\nu} k_j ={-} A^{{-}1}_{nj} \boldsymbol{\nabla} \delta \nu \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_j. \end{equation}

Inserting the above equation into (3.43) gives

(3.45)\begin{align} \delta_{\nu} \hat{\mathcal{L}} &={-}\rho \left[\dfrac{\partial{\delta \nu}}{\partial t} - A^{{-}1}_{nj} \boldsymbol{\nabla} \chi_j \dfrac{\partial{\chi_n}}{\partial t} \boldsymbol{\cdot} \boldsymbol{\nabla} \delta \nu\right] \nonumber\\ &={-}\rho \left[\frac{\partial{\delta \nu}}{\partial t} + \boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \delta \nu \right]={-}\rho \frac{\textrm{d}{\delta \nu}}{\textrm{d} t}. \end{align}

The above equation can be put to the form,

(3.46)\begin{equation} \delta_{\nu} \hat{\mathcal{L}} = \delta \nu \left[\frac{\partial{\rho}}{\partial t} + \boldsymbol{\nabla} \boldsymbol{\cdot} (\rho \boldsymbol{v})\right] -\frac{\partial{(\rho \delta \nu)}}{\partial t}- \boldsymbol{\nabla} \boldsymbol{\cdot} (\rho \boldsymbol{v} \delta \nu ). \end{equation}

This obviously leads to the continuity equation (2.3) and some boundary terms in space and time. The variational derivative with respect to $\rho$ 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 MHD as follows:

(3.47)\begin{equation} \left.\begin{gathered} \frac{\textrm{d} \nu}{\textrm{d} t} = \frac{1}{2} \boldsymbol{v}^2 - w, \frac{\partial{\rho}}{\partial t} + \boldsymbol{\nabla} \boldsymbol{\cdot} (\rho \boldsymbol{v} ) = 0, \\ \frac{\textrm{d} \sigma}{\textrm{d} t} = T,\quad \frac{\textrm{d} \alpha}{\textrm{d} t} = \frac{\boldsymbol{\nabla} \eta \boldsymbol{\cdot} \boldsymbol{J}}{\rho},\quad \frac{\textrm{d} \beta}{\textrm{d} t} ={-}\frac{\boldsymbol{\nabla} \chi \boldsymbol{\cdot} \boldsymbol{J}}{\rho} \end{gathered}\right\}\end{equation}

in which $\alpha ,\beta ,\sigma ,\boldsymbol {v}$ are functionals of $\chi ,\eta ,s,\nu$ as described above. It is easy to show as in (3.16) that those variational equations are equivalent to the physical equations.

3.6. Lagrangian density in explicit form

Let us put the Lagrangian density of (3.38) in a slightly more explicit form. First us look at the term $\boldsymbol {v}^2$,

(3.48)\begin{align} \boldsymbol{v}^2 &= A^{{-}1}_{ij}\boldsymbol{\nabla} \chi_i \frac{\partial \chi_j}{\partial t} A^{{-}1}_{mn}\boldsymbol{\nabla} \chi_m \frac{\partial \chi_n}{\partial t} \nonumber\\ &= A^{{-}1}_{ij} A^{{-}1}_{mn} A_{im} \frac{\partial \chi_j}{\partial t} \frac{\partial \chi_n}{\partial t} = A^{{-}1}_{jn} \frac{\partial \chi_j}{\partial t} \frac{\partial \chi_n}{\partial t}, \end{align}

where in the above we use (3.37) and (3.30). Next let us look at the expression

(3.49)\begin{align} \alpha_k [\chi_i,\nu] \frac{\partial{\chi_k}}{\partial t} &= A^{{-}1}_{kj} k_j \frac{\partial{\chi_k}}{\partial t}={-}\left(\frac{\partial \chi_j}{\partial t} + \boldsymbol{\nabla} \nu \boldsymbol{\cdot} \boldsymbol{\nabla} \chi_j\right)A^{{-}1}_{kj} \frac{\partial{\chi_k}}{\partial t} \nonumber\\ &={-}A^{{-}1}_{jk} \frac{\partial \chi_j}{\partial t} \frac{\partial \chi_k}{\partial t} -\frac{\partial \nu}{\partial \chi_m} \frac{\partial{\chi_m}}{\partial t}. \end{align}

Inserting (3.48) and (3.49) into (3.38) leads to a Lagrangian density of a more standard quadratic form,

(3.50) \begin{align} \hat{\mathcal{L}}[\chi_i,\nu,\rho] &= \rho \left[\frac{1}{2} A^{{-}1}_{jn} \frac{\partial \chi_j}{\partial t} \frac{\partial \chi_n}{\partial t}+\frac{\partial \nu}{\partial \chi_m} \frac{\partial{\chi_m}}{\partial t}-\frac{\partial{\nu}}{\partial t} -\varepsilon (\rho,\chi_3)\right]\nonumber\\ &\quad -\frac{1}{8 {\rm \pi}}(\boldsymbol{\nabla} \chi_1 \times \boldsymbol{\nabla} \chi_2)^2. \end{align}

We now define the metric $g_{jn} = A^{-1}_{jn}$ and obtain the geometrical Lagrangian,

(3.51)\begin{align} \hat{\mathcal{L}}[\chi_i,\nu,\rho] &= \rho \left[\frac{1}{2} g_{jn} \frac{\partial \chi_j}{\partial t} \frac{\partial \chi_n}{\partial t} +\frac{\partial \nu}{\partial \chi_m} \frac{\partial{\chi_m}}{\partial t}- \frac{\partial{\nu}}{\partial t} -\varepsilon (\rho,\chi_3)\right]\nonumber\\ &\quad \vphantom{\frac{\partial \chi_j}{\partial t}}-\frac{1}{8 {\rm \pi}}(\boldsymbol{\nabla} \chi_1 \times \boldsymbol{\nabla} \chi_2)^2. \end{align}

The Lagrangian is thus composed of a geometric kinetic term 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.

4.1. Lagrangian and Eulerian variations

The value of a function $f$ can be modified by evaluating it at a different point in space, the difference between the new and old values would be

(4.4)\begin{equation} f (\boldsymbol{x} + \boldsymbol{\xi}) - f (\boldsymbol{x}) = \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} f, \end{equation}

in which $\boldsymbol {x}$ is a coordinate vector and $\boldsymbol {\xi }$ is a displacement vector, the equality is correct to first order in $\boldsymbol {\xi }$. Alternatively we can modify the value of a function by changing it to a different function $f'$, in this case the difference between the new and old values would be

(4.5)\begin{equation} \delta f = f' (\boldsymbol{x}) - f (\boldsymbol{x}). \end{equation}

For a small $\delta f$ this just the standard variation of variational analysis or an Eulerian variation. Finally we can do both, in the last case the difference between the new and old values would be

(4.6)\begin{equation} {\rm \Delta} f = f' (\boldsymbol{x} + \boldsymbol{\xi}) - f (\boldsymbol{x}), \end{equation}

hence

(4.7)\begin{equation} {\rm \Delta} f = f' (\boldsymbol{x} + \boldsymbol{\xi}) - f (\boldsymbol{x} + \boldsymbol{\xi}) + f (\boldsymbol{x} + \boldsymbol{\xi}) - f (\boldsymbol{x}). \end{equation}

Keeping only first-order terms we obtain

(4.8)\begin{equation} {\rm \Delta} f = \delta f + \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} f \Rightarrow \delta f = {\rm \Delta} f - \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} f, \end{equation}

in which ${\rm \Delta}$ is a Lagrangian variation.

Now suppose that a specific function is connected to a fluid element in such a way that its value in space is determined only by fluid element location. And suppose that the fluid element is displaced as dictated by the flow. Such a function would be denoted a label of the flow and its material derivative would vanish. Moreover, for a label

(4.9)\begin{equation} f' (\boldsymbol{x} + \boldsymbol{\xi}) = f (\boldsymbol{x}) \Rightarrow {\rm \Delta} f = 0, \end{equation}

in order to change the value of a label in a certain point in space the fluid element must be displaced and another (with a different label value) must take its place. If follows from (4.8) that for a label

(4.10)\begin{equation} \delta f ={-} \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} f. \end{equation}

Now suppose we have a set of three labels $\tilde \chi _i$ such that

(4.11)\begin{equation} \delta \tilde \chi_i={-} \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} \tilde \chi_i ={-}\xi_k \frac{\partial \tilde \chi_i}{\partial x_k}, \end{equation}

in which we use the Einstein summation convention and $x_k$ are Cartesian coordinates. The inverse of the matrix ${\partial \tilde \chi _i}/{\partial x_k}$ is ${\partial x_k}/{\partial \tilde \chi _i}$ as

(4.12)\begin{equation} \frac{\partial \tilde \chi_i}{\partial x_k} \frac{\partial x_j}{\partial \tilde \chi_i} = \delta_k^j, \end{equation}

where $\delta _k^j$ is a Kronecker delta. It thus follows that one can calculate the displacement vector $\boldsymbol {\xi }$ as follows:

(4.13)\begin{equation} \xi_k ={-} \frac{\partial x_k}{\partial \tilde \chi_i} \delta \tilde \chi_i \Rightarrow \boldsymbol{\xi} ={-}\frac{\partial \boldsymbol{r}}{\partial \tilde \chi_i} \delta \tilde \chi_i. \end{equation}

4.2. Noether current for label symmetries

We now study the form of the Noether current equation (4.3) for the case of label symmetry transformations. It is clear from (3.3) that $\chi ,\eta$ can be taken to be labels. Hence we can write the conserved Noether current defined in (4.3) as

(4.14)\begin{equation} \delta J ={-} \int \textrm{d}^3 x \rho \left[ {\rm \Delta} \nu - \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} \nu - \alpha \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} \chi - \beta \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} \eta \right], \end{equation}

or using (3.6) as

(4.15)\begin{equation} \delta J = \int \textrm{d}^3 x \rho \left[\boldsymbol{\xi} \cdot (\boldsymbol{v} - \sigma \boldsymbol{\nabla} s) - {\rm \Delta} \nu \right]. \end{equation}

We use the topological velocity defined in (3.21) to get

(4.16)\begin{equation} \boldsymbol{v}_t = \boldsymbol{\nabla} \nu + \alpha \boldsymbol{\nabla} \chi + \beta \boldsymbol{\nabla} \eta, \end{equation}

and write

(4.17)\begin{equation} \delta J = \int \textrm{d}^3 x \rho \left[\boldsymbol{\xi} \cdot \boldsymbol{v}_t - {\rm \Delta} \nu \right]. \end{equation}

Suppose now that we are considering label symmetry transformations with the infinitesimal form $\tilde \chi _i + \delta \tilde \chi _i$, this type of transformation will induce a transformation on other functions ($\nu$ as an example) which could be thought of as functions of labels, a transformation of the form

(4.18)\begin{equation} \delta \nu= \nu (\tilde \chi_i + \delta \tilde \chi_i) - \nu (\tilde \chi_i ) = \delta \tilde \chi_i \partial_{\tilde \chi_i} \nu ={-} \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} \tilde \chi_i \partial_{\tilde \chi_i} \nu ={-} \boldsymbol{\xi} \boldsymbol{\cdot} \boldsymbol{\nabla} \nu. \end{equation}

Hence by (4.8) we have

(4.19)\begin{equation} {\rm \Delta} \nu = 0. \end{equation}

It follows that for an induced infinitesimal label transformation any function will transform as a label. From (4.19) it follows that the Noether current will take the following form for a symmetry label transformation:

(4.20)\begin{equation} \delta J = \int \textrm{d}^3 x \rho \boldsymbol{\xi} \cdot \boldsymbol{v}_t. \end{equation}

This Noether current form is identical to (47) of Yahalom (Reference Yahalom and Dobrev2017d) and (14) of Yahalom (Reference Yahalom2019b), which were derived from a Lagrangian variational principle. We note, however, that this form is limited to the case of label transformations and the general form given in (4.2) allows us to exploit larger symmetry groups. Next we will study some symmetry transformations of the action $A$, in order to do this we shall first introduce the load and metage quantities.

6.1. Metage translations

In what follows we consider the transformation (see also (5.6))

(6.9)\begin{equation} \tilde \chi = \chi, \tilde \eta = \eta, \tilde \mu = \mu + a (\chi,\eta). \end{equation}

Hence $a$ is a label displacement which may be different for each magnetic field line, as the field line is closed one need not worry about edge difficulties. This transformation satisfies trivially the conditions (6.1), (6.3). If we take the infinitesimal symmetry transformation $\delta \mu = a, \delta \chi =\delta \eta =0$ we can calculate the associated fluid element displacement with this relabeling using (4.13) and (5.8), giving

(6.10)\begin{equation} \boldsymbol{\xi} ={-}\frac{\partial \boldsymbol{r}}{\partial \mu}\delta \mu ={-}\delta \mu \frac{\boldsymbol{B}}{\rho}. \end{equation}

Inserting (6.10) into (4.20) we obtain the conservation law,

(6.11)\begin{equation} \delta J =\int \textrm{d}^3 x \rho \boldsymbol{v}_t \cdot \boldsymbol{\xi} ={-}\int \textrm{d}^3 x \delta \mu \boldsymbol{v}_t \cdot \boldsymbol{B}. \end{equation}

In the simplest case we may take $\delta \mu$ to be a small constant, hence

(6.12)\begin{equation} \delta J ={-} \delta \mu \int \textrm{d}^3 x \boldsymbol{v}_t \cdot \boldsymbol{B} ={-} \delta \mu H_{CNB}, \end{equation}

where $H_{CNB}$ is the non-barotropic global cross-helicity (Webb et al. Reference Webb, Dasgupta, McKenzie, Hu and Zank2014a; Yahalom Reference Yahalom2017b,Reference Yahalomc) defined as

(6.13)\begin{equation} H_{CNB} \equiv \int \textrm{d}^3 x \boldsymbol{v}_t \cdot \boldsymbol{B}. \end{equation}

We thus obtain the conservation of non-barotropic cross-helicity using Noether's theorem and the symmetry group of metage translations. Of course one can perform a different translation on each magnetic field line, in this case one obtains

(6.14)\begin{equation} \delta J ={-}\int \textrm{d}^3 x \delta \mu \boldsymbol{v}_t \cdot \boldsymbol{B} ={-}\int \textrm{d} \chi \,\textrm{d} \eta \delta \mu \oint_{\chi,\eta} \textrm{d}\mu \rho^{{-}1} \boldsymbol{v}_t \cdot \boldsymbol{B}. \end{equation}

Now since $\delta \mu$ is an arbitrary (small) function of $\chi ,\eta$ it follows that

(6.15)\begin{equation} I = \oint_{\chi,\eta} \textrm{d}\mu \rho^{{-}1} \boldsymbol{v}_t \cdot \boldsymbol{B} \end{equation}

is a conserved quantity for each magnetic field line. Along a magnetic field line the following equations hold:

(6.16)\begin{equation} \textrm{d}\mu = \boldsymbol{\nabla} \mu \boldsymbol{\cdot} \textrm{d} \boldsymbol{r} = \boldsymbol{\nabla} \mu \boldsymbol{\cdot} \hat{B} \,\textrm{d} r = \frac{\rho}{B} \textrm{d} r. \end{equation}

In the above, $\hat {B}$ is a unit vector in the magnetic field direction and (5.8) is used. Inserting (6.16) into (6.15) we obtain

(6.17)\begin{equation} I = \oint_{\chi,\eta} \textrm{d} r \boldsymbol{v}_t \cdot \hat B = \oint_{\chi,\eta} \textrm{d} \boldsymbol{r} \cdot \boldsymbol{v}_t, \end{equation}

which is just the circulation of the topological velocity along the magnetic field lines. This quantity can be written in terms of the generalized Clebsch representation of the velocity equation (3.6) as

(6.18)\begin{equation} I = \oint_{\chi,\eta} \textrm{d} \boldsymbol{r} \cdot \boldsymbol{v}_t = \oint_{\chi,\eta} \textrm{d} \boldsymbol{r} \cdot \boldsymbol{\nabla} \nu = [\nu], \end{equation}

where $[\nu ]$ is the discontinuity of $\nu$. This was shown to be equal to the amount of non-barotropic cross-helicity per unit of magnetic flux in (3.23) (Yahalom Reference Yahalom2017b,Reference Yahalomc),

(6.19)\begin{equation} I=[\nu]= \frac{\textrm{d} H_{CNB}}{\textrm{d} \varPhi}. \end{equation}

6.2. Transformations of magnetic surfaces

Consider the following transformations:

(6.20a–c)\begin{equation} \tilde{\eta} = \eta + \delta \eta(\chi,\eta),\quad \tilde{\chi} = \chi + \delta \chi(\chi,\eta),\quad \tilde{\mu} = \mu, \end{equation}

in which $\delta \eta ,\delta \chi$ are considered small in some sense. Inserting the above quantities into (6.3) and keeping only first-order terms we arrive at

(6.21)\begin{equation} \partial_{\eta} \delta \eta + \partial_{\chi} \delta \chi = 0. \end{equation}

This equation can be solved as follows:

(6.22a,b)\begin{equation} \delta \eta = \partial_{\chi} \delta f,\quad \delta \chi ={-}\partial_{\eta} \delta f, \end{equation}

in which $\delta f= \delta f (\chi ,\eta )$ is an arbitrary small function. In this case we obtain a particle displacement of the form,

(6.23)\begin{align} \boldsymbol{\xi} &={-}\frac{\partial \boldsymbol{r}}{\partial \chi}\delta \chi -\frac{\partial \boldsymbol{r}}{\partial \eta}\delta \eta ={-}\frac{1}{\rho} \left( \boldsymbol{\nabla} \eta \times \boldsymbol{\nabla} \mu \delta \chi + \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \chi \delta \eta \right) \nonumber\\ &= \frac{\boldsymbol{\nabla} \mu}{\rho} \times ( \boldsymbol{\nabla} \eta \delta \chi - \boldsymbol{\nabla} \chi \delta \eta ). \end{align}

A special case that satisfies (6.21) is the case of a constant $\delta \chi$ and $\delta \eta$, those two independent displacements lead to the following two new topological conservation laws,

(6.24)\begin{equation} \left.\begin{gathered} \delta J_\chi = \delta \chi \int \textrm{d}^3 x \boldsymbol{v} _t \boldsymbol{\cdot} \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta= \delta \chi H_{CNB \chi}, \\ \delta J_\eta = \delta \eta \int \textrm{d}^3 x \boldsymbol{v} _t \boldsymbol{\cdot} \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu = \delta \eta H_{CNB \eta}, \end{gathered}\right\}\end{equation}

where the new non-barotropic global cross-helicities are defined as

(6.25a,b)\begin{equation} H_{CNB \chi} \equiv \int \textrm{d}^3 x \boldsymbol{v} _t \boldsymbol{\cdot} \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta,\quad H_{CNB \eta} \equiv \int \textrm{d}^3 x \boldsymbol{v} _t \boldsymbol{\cdot} \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu. \end{equation}

We will find it useful to introduce the abstract ‘magnetic fields’ as follows:

(6.26a,b)\begin{equation} \boldsymbol{B}_\chi \equiv \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta,\quad \boldsymbol{B}_\eta \equiv \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu, \end{equation}

in terms of which we obtain the new helicities in a more conventional form, i.e.

(6.27a,b)\begin{equation} H_{CNB \chi} = \int \textrm{d}^3 x \boldsymbol{v} _t \cdot \boldsymbol{B}_\chi,\quad H_{CNB \eta} = \int \textrm{d}^3 x \boldsymbol{v} _t \cdot \boldsymbol{B}_\eta. \end{equation}

It is more plausible that those symmetries and conservation laws hold for magnetic field lines which lie on topological tori. In this case $\eta$ is non-single-valued (Yahalom & Lynden-Bell Reference Yahalom and Lynden-Bell2008) and thus the translation in this direction resembles moving fluid elements along closed loops. Both those helicities suffer a topological interpretation in terms of the knottiness of the abstract magnetic field lines and the flow lines. Finally we remark that for barotropic MHD $\boldsymbol {v}_t$ can be replaced with $\boldsymbol {v}$.

7.1. Direct derivation of the constancy of non-barotropic cross-helicity

Taking the temporal derivative of the non-barotropic cross-helicity given in (6.13) we obtain,

(7.1)\begin{equation} \frac{\textrm{d} H_{CNB}}{\textrm{d} t} = \int \textrm{d}^3 x \left[ \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B} + \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B} \right], \end{equation}

where ${\textrm {d}}/{\textrm {d} t}$ is an ordinary temporal derivative, and we use the notation, $\partial _t \equiv {\partial }/{\partial t}$. Using (2.1) it follows that:

(7.2)\begin{equation} \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B} = \boldsymbol{v}_t \boldsymbol{\cdot} \boldsymbol{\nabla} \times (\boldsymbol{v} \times \boldsymbol{B})= \boldsymbol{\nabla} \boldsymbol{\cdot} \left( (\boldsymbol{v} \times \boldsymbol{B}) \times \boldsymbol{v}_t \right) + (\boldsymbol{v} \times \boldsymbol{B}) \cdot \boldsymbol{\omega}_t, \end{equation}

in which we have used a standard identity of vector analysis and the definition,

(7.3)\begin{equation} \boldsymbol{\omega}_t = \boldsymbol{\nabla} \times \boldsymbol{v}_t = \boldsymbol{\omega} - \boldsymbol{\nabla} \sigma \times \boldsymbol{\nabla} s, \end{equation}

where $\boldsymbol {v}_t$ is defined in (3.21). Next we calculate

(7.4)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B} = \boldsymbol{B} \cdot \partial_t \left(\boldsymbol{v} - \sigma \boldsymbol{\nabla} s \right) = \boldsymbol{B} \cdot \left( \partial_t \boldsymbol{v} - \partial_t \sigma \boldsymbol{\nabla} s - \sigma \boldsymbol{\nabla} \partial_t s \right). \end{equation}

Taking into account Euler's equations (2.4) and the standard thermodynamic identities of (2.6) we have,

(7.5)\begin{align} \partial_t \boldsymbol{v} &={-}(\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla})\boldsymbol{v} - \frac{1}{\rho}\boldsymbol{\nabla} p (\rho,s) + \frac{(\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B}}{4 {\rm \pi}\rho} \nonumber\\ &= \boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\frac{1}{2} \boldsymbol{v}^2\right) - \boldsymbol{\nabla} w + T \boldsymbol{\nabla} s + \frac{(\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B}}{4 {\rm \pi}\rho}. \end{align}

Hence we have

(7.6)\begin{equation} \boldsymbol{B} \cdot \partial_t \boldsymbol{v} = \boldsymbol{B} \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\tfrac{1}{2} \boldsymbol{v}^2+w\right) + T \boldsymbol{\nabla} s \right). \end{equation}

Taking into account (3.13) it follows that:

(7.7)\begin{equation} - \partial_t \sigma \boldsymbol{\nabla} s = (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \sigma - T) \boldsymbol{\nabla} s. \end{equation}

And taking into account (2.5) it follows that:

(7.8)\begin{equation} - \sigma \boldsymbol{\nabla} \partial_t s = \sigma \boldsymbol{\nabla} (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s). \end{equation}

Inserting (7.6), (7.7) and (7.8) into (7.4) it follows that:

(7.9)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B} = \boldsymbol{B} \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\tfrac{1}{2} \boldsymbol{v}^2+w\right) + (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \sigma) \boldsymbol{\nabla} s + \sigma \boldsymbol{\nabla} (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s) \right), \end{equation}

hence

(7.10)\begin{align} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B} &= \boldsymbol{B} \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} + \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) + (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \sigma) \boldsymbol{\nabla} s - (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s) \boldsymbol{\nabla} \sigma \right) \nonumber\\ &=\boldsymbol{B} \cdot \left( \boldsymbol{v} \times \boldsymbol{\omega} + \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) + (\boldsymbol{\nabla} \sigma \times \boldsymbol{\nabla} s) \times \boldsymbol{v} \right) \nonumber\\ &= \boldsymbol{B} \cdot \left( \boldsymbol{v} \times \boldsymbol{\omega}_t + \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) \right). \end{align}

Combining (7.2) with (7.10) we arrive at the result,

(7.11)\begin{align} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B} + \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B} &= \boldsymbol{\nabla} \boldsymbol{\cdot} \left( (\boldsymbol{v} \times \boldsymbol{B}) \times \boldsymbol{v}_t \right) + \boldsymbol{B} \boldsymbol{\cdot} \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)- \tfrac{1}{2} \boldsymbol{v}^2-w\right) \nonumber\\ &= \boldsymbol{\nabla} \boldsymbol{\cdot} \left[ (\boldsymbol{v} \times \boldsymbol{B}) \times \boldsymbol{v}_t + \boldsymbol{B} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) \right], \end{align}

in which we take into account (2.2). Inserting (7.11) into (7.1) and using Gauss’ theorem we obtain a surface integral,

(7.12)\begin{equation} \frac{\textrm{d} H_{CNB}}{\textrm{d} t} = \oint \textrm{d} \boldsymbol{S} \cdot \left[(\boldsymbol{v} \times \boldsymbol{B}) \times \boldsymbol{v}_t + \boldsymbol{B} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\frac{1}{2} \boldsymbol{v}^2-w\right) \right].\end{equation}

The surface integral encapsulates the volume for which the non-barotropic cross-helicity is calculated. If the surface is taken at infinity the magnetic fields vanish and thus

(7.13)\begin{equation} \frac{\textrm{d} H_{CNB}}{\textrm{d} t} = 0, \end{equation}

which means that $H_{CNB}$ is a constant of motion. We notice the complexity of the direct derivation with respect to the elegance and simplicity of the Noether theorem approach. However, obtaining the same result using different methods strengthens our confidence that no mathematical error was accidentally introduced.

7.2. Direct derivation of the constancy of non-barotropic $\chi$ cross-helicity

Taking the temporal derivative of the non-barotropic $\chi$ cross-helicity given in (6.27a,b) we obtain

(7.14)\begin{equation} \frac{\textrm{d} H_{CNB\chi}}{\textrm{d} t} = \int \textrm{d}^3 x \left[\partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\chi + \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B}_\chi \right]. \end{equation}

Let us calculate $\partial _t \boldsymbol {B}_\chi$ where $\boldsymbol {B}_\chi$ is defined in (6.26a,b):

(7.15)\begin{equation} \boldsymbol{B}_\chi = \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta. \end{equation}

It follows that

(7.16)\begin{equation} \partial_t \boldsymbol{B}_\chi = \boldsymbol{\nabla} \partial_t \mu \times \boldsymbol{\nabla} \eta + \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \partial_t \eta. \end{equation}

Using (3.3) and (5.7) we obtain

(7.17)\begin{align} \partial_t \boldsymbol{B}_\chi &= \boldsymbol{\nabla} (-\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \mu) \times \boldsymbol{\nabla} \eta+ \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} (-\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \eta) \nonumber\\ &= \boldsymbol{\nabla} \times \left(\boldsymbol{\nabla} \mu (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \eta) - \boldsymbol{\nabla} \eta (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \mu)\right) \nonumber\\ &= \boldsymbol{\nabla} \times \left(\boldsymbol{v} \times (\boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta) \right) = \boldsymbol{\nabla} \times \left(\boldsymbol{v} \times \boldsymbol{B}_\chi \right), \end{align}

in which we used standard vector analysis identities. It thus follows that

(7.18)\begin{equation} \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B}_\chi = \boldsymbol{v}_t \boldsymbol{\cdot} \boldsymbol{\nabla} \times (\boldsymbol{v} \times \boldsymbol{B}_\chi)= \boldsymbol{\nabla} \boldsymbol{\cdot} \left( (\boldsymbol{v} \times \boldsymbol{B}_\chi) \times \boldsymbol{v}_t \right) + (\boldsymbol{v} \times \boldsymbol{B}_\chi) \cdot \boldsymbol{\omega}_t. \end{equation}

Next we calculate

(7.19)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\chi = \boldsymbol{B}_\chi \cdot \partial_t \left(\boldsymbol{v} - \sigma \boldsymbol{\nabla} s \right)= \boldsymbol{B}_\chi \cdot \left(\partial_t \boldsymbol{v} - \partial_t \sigma \boldsymbol{\nabla} s - \sigma \boldsymbol{\nabla} \partial_t s \right). \end{equation}

Taking into account (7.5),

(7.20)\begin{equation} \boldsymbol{B}_\chi \cdot \partial_t \boldsymbol{v} = \boldsymbol{B}_\chi \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\frac{1}{2} \boldsymbol{v}^2+w\right) + T \boldsymbol{\nabla} s \right) + \boldsymbol{B}_\chi \cdot \frac{1}{\rho}\boldsymbol{J} \times \boldsymbol{B}, \end{equation}

in which the current density is given by

(7.21)\begin{equation} \boldsymbol{J} = \frac{\boldsymbol{\nabla} \times \boldsymbol{B}}{4 {\rm \pi}} \quad \Rightarrow \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{J} = 0. \end{equation}

Now,

(7.22)\begin{equation} \boldsymbol{B}_\chi \cdot \frac{1}{\rho}\boldsymbol{J} \times \boldsymbol{B} = \frac{1}{\rho} \boldsymbol{J} \cdot \boldsymbol{B} \times \boldsymbol{B}_\chi, \end{equation}

however,

(7.23)\begin{equation} \boldsymbol{B} \times \boldsymbol{B}_\chi = \boldsymbol{B} \times (\boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta) = \boldsymbol{\nabla} \mu (\boldsymbol{B} \boldsymbol{\cdot} \boldsymbol{\nabla} \eta) - \boldsymbol{\nabla} \eta (\boldsymbol{B} \boldsymbol{\cdot} \boldsymbol{\nabla} \mu) ={-} \rho \boldsymbol{\nabla} \eta. \end{equation}

It thus follows that

(7.24)\begin{equation} \boldsymbol{B}_\chi \cdot \frac{1}{\rho}\boldsymbol{J} \times \boldsymbol{B} ={-} \boldsymbol{J} \boldsymbol{\cdot} \boldsymbol{\nabla} \eta ={-} \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \eta), \end{equation}

in which we used (3.4) and (5.8). Inserting (7.24) into (7.20) will yield

(7.25)\begin{equation} \boldsymbol{B}_\chi \cdot \partial_t \boldsymbol{v} = \boldsymbol{B}_\chi \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\tfrac{1}{2} \boldsymbol{v}^2+w\right) + T \boldsymbol{\nabla} s \right) - \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \eta). \end{equation}

Inserting (7.25), (7.7) and (7.8) into (7.19) it follows that

(7.26)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\chi = \boldsymbol{B}_\chi \cdot \left( \boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\tfrac{1}{2} \boldsymbol{v}^2+w\right) + (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \sigma) \boldsymbol{\nabla} s + \sigma \boldsymbol{\nabla} (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s) \right)- \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \eta)\end{equation}

hence

(7.27)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\chi= \boldsymbol{B}_\chi \cdot \left( \boldsymbol{v} \times \boldsymbol{\omega}_t + \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) \right)- \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \eta).\end{equation}

Combining (7.18) with (7.27) we arrive at the result

(7.28)\begin{align} & \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\chi + \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B}_\chi \nonumber\\ &\quad =\boldsymbol{\nabla} \boldsymbol{\cdot} \left( (\boldsymbol{v} \times \boldsymbol{B}_\chi) \times \boldsymbol{v}_t - \boldsymbol{J} \eta \right) + \boldsymbol{B}_\chi \boldsymbol{\cdot} \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) \nonumber\\ &\quad =\boldsymbol{\nabla} \boldsymbol{\cdot} \left[ (\boldsymbol{v} \times \boldsymbol{B}_\chi) \times \boldsymbol{v}_t + \boldsymbol{B}_\chi \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) - \boldsymbol{J} \eta \right], \end{align}

in which we take into account (7.15). Inserting (7.28) into (7.14) and using Gauss’ theorem we obtain a surface integral,

(7.29)\begin{align} \frac{\textrm{d} H_{CNB \chi}}{\textrm{d} t} &= \oint \textrm{d} \boldsymbol{S} \cdot \left[ (\boldsymbol{v} \times \boldsymbol{B}_\chi) \times \boldsymbol{v}_t + \boldsymbol{B}_\chi \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\frac{1}{2} \boldsymbol{v}^2-w\right) - \boldsymbol{J} \eta \right] \nonumber\\ &\quad -\int \textrm{d} \boldsymbol{\varSigma} \cdot \boldsymbol{J} [\eta]. \end{align}

The surface integral encapsulates the volume for which the $\chi$ non-barotropic cross-helicity is calculated and an additional surface integral is performed along the cut of $\eta$, in case that $\eta$ is not single valued (see (5.18)). If the surface is taken at infinity the magnetic fields and current densities vanish and thus

(7.30)\begin{equation} \frac{\textrm{d} H_{CNB \chi}}{\textrm{d} t} ={-} \int \textrm{d} \boldsymbol{\varSigma} \cdot \boldsymbol{J} [\eta]; \end{equation}

hence for spherical topologies of magnetic field lines or for a current density $\boldsymbol {J}$ parallel to the cut we obtain

(7.31)\begin{equation} \frac{\textrm{d} H_{CNB \chi}}{\textrm{d} t} = 0, \end{equation}

which means that $H_{CNB \chi }$ is a constant of motion. We notice the complexity of the direct derivation with respect to the elegance and simplicity of the Noether theorem approach. However, obtaining the same result using different methods strengthens our confidence that no mathematical error was accidentally introduced.

7.3. Direct derivation of the constancy of non-barotropic $\eta$ cross-helicity

Taking the temporal derivative of the non-barotropic $\eta$ cross-helicity given in (6.27a,b) we obtain

(7.32)\begin{equation} \frac{\textrm{d} H_{CNB\eta}}{\textrm{d} t} = \int \textrm{d}^3 x \left[ \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\eta + \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B}_\eta \right]. \end{equation}

Let us calculate $\partial _t \boldsymbol {B}_\eta$ where $\boldsymbol {B}_\eta$ is defined in (6.26a,b):

(7.33)\begin{equation} \boldsymbol{B}_\eta = \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu; \end{equation}

it follows that

(7.34)\begin{equation} \partial_t \boldsymbol{B}_\eta = \boldsymbol{\nabla} \partial_t \chi \times \boldsymbol{\nabla} \mu + \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \partial_t \mu. \end{equation}

Using (3.3) and (5.7) we obtain

(7.35)\begin{align} \partial_t \boldsymbol{B}_\eta &= \boldsymbol{\nabla} (-\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \chi) \times \boldsymbol{\nabla} \mu+ \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} (-\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \mu) \nonumber\\ &= \boldsymbol{\nabla} \times \left(\boldsymbol{\nabla} \chi (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \mu) - \boldsymbol{\nabla} \mu (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \chi)\right) \nonumber\\ &= \boldsymbol{\nabla} \times \left(\boldsymbol{v} \times (\boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu) \right) = \boldsymbol{\nabla} \times \left(\boldsymbol{v} \times \boldsymbol{B}_\eta \right), \end{align}

in which we used standard vector analysis identities. It thus follows that

(7.36)\begin{equation} \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B}_\eta = \boldsymbol{v}_t \boldsymbol{\cdot} \boldsymbol{\nabla} \times (\boldsymbol{v} \times \boldsymbol{B}_\eta)= \boldsymbol{\nabla} \boldsymbol{\cdot} \left( (\boldsymbol{v} \times \boldsymbol{B}_\eta) \times \boldsymbol{v}_t \right) + (\boldsymbol{v} \times \boldsymbol{B}_\eta) \cdot \boldsymbol{\omega}_t .\end{equation}

Next we calculate

(7.37)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\eta = \boldsymbol{B}_\eta \cdot \partial_t \left(\boldsymbol{v} - \sigma \boldsymbol{\nabla} s \right)= \boldsymbol{B}_\eta \cdot \left( \partial_t \boldsymbol{v} - \partial_t \sigma \boldsymbol{\nabla} s - \sigma \boldsymbol{\nabla} \partial_t s \right). \end{equation}

Taking into account (7.5) we have

(7.38)\begin{equation} \boldsymbol{B}_\eta \cdot \partial_t \boldsymbol{v} = \boldsymbol{B}_\eta \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\frac{1}{2} \boldsymbol{v}^2+w\right) + T \boldsymbol{\nabla} s \right) + \boldsymbol{B}_\eta \cdot \frac{1}{\rho}\boldsymbol{J} \times \boldsymbol{B}. \end{equation}

Now,

(7.39)\begin{equation} \boldsymbol{B}_\eta \cdot \frac{1}{\rho}\boldsymbol{J} \times \boldsymbol{B} = \frac{1}{\rho} \boldsymbol{J} \cdot \boldsymbol{B} \times \boldsymbol{B}_\eta, \end{equation}

however,

(7.40)\begin{equation} \boldsymbol{B} \times \boldsymbol{B}_\eta = \boldsymbol{B} \times (\boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu) = \boldsymbol{\nabla} \chi (\boldsymbol{B} \boldsymbol{\cdot} \boldsymbol{\nabla} \mu) - \boldsymbol{\nabla} \mu (\boldsymbol{B} \boldsymbol{\cdot} \boldsymbol{\nabla} \chi) = \rho \boldsymbol{\nabla} \chi; \end{equation}

it thus follows that

(7.41)\begin{equation} \boldsymbol{B}_\eta \cdot \frac{1}{\rho}\boldsymbol{J} \times \boldsymbol{B} = \boldsymbol{J} \boldsymbol{\cdot} \boldsymbol{\nabla} \chi = \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \chi), \end{equation}

in which we used (3.4) and (5.8). Inserting (7.41) into (7.38) will yield

(7.42)\begin{equation} \boldsymbol{B}_\eta \cdot \partial_t \boldsymbol{v} = \boldsymbol{B}_\eta \cdot \left(\boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\tfrac{1}{2} \boldsymbol{v}^2+w\right) + T \boldsymbol{\nabla} s \right) + \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \chi). \end{equation}

Inserting (7.42), (7.7) and (7.8) into (7.37) it follows that:

(7.43)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\eta = \boldsymbol{B}_\eta \cdot \left( \boldsymbol{v} \times \boldsymbol{\omega} - \boldsymbol{\nabla} \left(\tfrac{1}{2} \boldsymbol{v}^2+w\right) + (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \sigma) \boldsymbol{\nabla} s + \sigma \boldsymbol{\nabla} (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s) \right)+ \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \chi), \end{equation}

hence

(7.44)\begin{equation} \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\eta= \boldsymbol{B}_\eta \cdot \left( \boldsymbol{v} \times \boldsymbol{\omega}_t + \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) \right)+ \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{J} \chi).\end{equation}

Combining (7.36) with (7.44) we arrive at the result,

(7.45)\begin{align} & \partial_t \boldsymbol{v}_t \cdot \boldsymbol{B}_\eta + \boldsymbol{v}_t \cdot \partial_t \boldsymbol{B}_\eta \nonumber\\ &\quad =\boldsymbol{\nabla} \boldsymbol{\cdot} \left( (\boldsymbol{v} \times \boldsymbol{B}_\eta) \times \boldsymbol{v}_t + \boldsymbol{J} \chi \right) + \boldsymbol{B}_\eta \boldsymbol{\cdot} \boldsymbol{\nabla} \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) \nonumber\\ &\quad =\boldsymbol{\nabla} \boldsymbol{\cdot} \left[ (\boldsymbol{v} \times \boldsymbol{B}_\eta) \times \boldsymbol{v}_t + \boldsymbol{B}_\eta \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) + \boldsymbol{J} \chi \right], \end{align}

in which we take into account (7.33). Inserting (7.45) into (7.32) and using Gauss’ theorem we obtain a surface integral,

(7.46)\begin{equation} \frac{\textrm{d} H_{CNB \eta}}{\textrm{d} t} = \oint \textrm{d} \boldsymbol{S} \cdot \left[ (\boldsymbol{v} \times \boldsymbol{B}_\eta) \times \boldsymbol{v}_t + \boldsymbol{B}_\eta \left(\sigma (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} s)-\tfrac{1}{2} \boldsymbol{v}^2-w\right) + \boldsymbol{J} \chi \right]. \end{equation}

The surface integral encapsulates the volume for which the $\eta$ non-barotropic cross-helicity is calculated. If the surface is taken at infinity the magnetic fields and current densities vanish and thus

(7.47)\begin{equation} \frac{\textrm{d} H_{CNB \eta}}{\textrm{d} t} = 0, \end{equation}

which means that $H_{CNB \eta }$ is a constant of motion. We notice the complexity of the direct derivation with respect to the elegance and simplicity of the Noether theorem approach. However, obtaining the same result using different methods strengthens our confidence that no mathematical error was accidentally introduced.

8.1. Bounds and constraints

In his important review paper ‘Physics of magnetically confined plasmas’, Boozer (Reference Boozer2004) states the following. ‘A spiky current profile causes a rapid dissipation of energy relative to magnetic helicity. If the evolution of a magnetic field is rapid, then it must be at constant helicity.’ Usually topological conservation laws are used in order to deduce lower bounds on the ‘energy’ of the flow. Those bounds are only approximate in non-ideal flows but due to their topological nature simulations show that they are approximately conserved even when the ‘energy’ is not. For example, it is easy to show that the ‘energy’ is bounded from below by the non-barotropic cross-helicity as follows (see Yahalom Reference Yahalom2019a):

(8.1)\begin{gather} \left| H_{CNB} \right|= \left|\int \boldsymbol{B}\cdot \boldsymbol{v}_t\,\textrm{d}^{3} x \right|\leq \tfrac{1}{2}\int \left( \boldsymbol{B}^2 + \boldsymbol{v}_t^2 \right) \textrm{d}^{3} x, \end{gather}

(8.2)\begin{gather}\left| H_{CNB} \right|= \left| \int \boldsymbol{B}\cdot \boldsymbol{v}_t\,\textrm{d}^{3} x \right|\leq \sqrt{\int \boldsymbol{v}_t^2 \,\textrm{d}^{3} x}\sqrt{\int \boldsymbol{B}^2 \,\textrm{d}^{3} x}, \end{gather}

where the second equation is a result of the Cauchy–Schwartz inequality. In this sense a configuration with a highly complicated topology is more stable since its energy is bounded from below. It is a simple thing to show that similar bounds occur also for the $\chi$ and $\eta$ helicities,

(8.3)\begin{gather} \left|H_{CNB \chi} \right|= \left|\int \boldsymbol{B}_\chi\cdot \boldsymbol{v}_t\,\textrm{d}^{3} x \right| \leq \tfrac{1}{2}\int \left( \boldsymbol{B}_\chi^2 + \boldsymbol{v}_t^2 \right) \textrm{d}^{3} x, \end{gather}

(8.4)\begin{gather}\left|H_{CNB \chi} \right|= \left|\int \boldsymbol{B}_\chi\cdot \boldsymbol{v}_t\,\textrm{d}^{3} x \right| \leq \sqrt{\int \boldsymbol{v}_t^2 \,\textrm{d}^{3} x}\sqrt{\int \boldsymbol{B}_\chi^2 \,\textrm{d}^{3} x}, \end{gather}

(8.5)\begin{gather}\left| H_{CNB \eta} \right|= \left|\int \boldsymbol{B}_\eta\cdot \boldsymbol{v}_t\,\textrm{d}^{3} x \right|\leq \tfrac{1}{2}\int \left( \boldsymbol{B}_\eta^2 + \boldsymbol{v}_t^2 \right) \textrm{d}^{3} x, \end{gather}

(8.6)\begin{gather}\left|H_{CNB \eta} \right|= \left|\int \boldsymbol{B}_\eta\cdot \boldsymbol{v}_t\,\textrm{d}^{3} x \right| \leq \sqrt{\int \boldsymbol{v}_t^2 \,\textrm{d}^{3} x}\sqrt{\int \boldsymbol{B}_\eta^2 \,\textrm{d}^{3} x}. \end{gather}

Hence the kinetic energy is bounded by three different bounds and so it the ‘total’ energy. The importance of each of those bounds is dependent on the flow.

8.2. A helical stratified magnetic field

8.2.1. The magnetic field and related labels

Consider a magnetohydrodynamic flow of uniform density $\rho$. Furthermore, assume that the flow contains a helical stratified magnetic field,

(8.7)\begin{equation} \boldsymbol{B} = \left\{\begin{array}{@{}ll} 2 B_{\bot}\left(1-\displaystyle\dfrac{R}{a}\right)\hat{\phi} + B_{z0}\hat{z} & R < a, \\ 0 & R > a, \end{array}\right. \end{equation}

in which $R,\phi ,z$ are the standard cylindrical coordinates, $\hat {R},\hat {\phi },\hat {z}$ are the corresponding unit vectors and $B_{z0},B_{\bot }$ are constants. The magnetic field is contained in a cylinder of radius $a$ and is independent of $z$. Furthermore, we assume that the planes $z=0$ and $z=L$ can be identified such that a topological torus is created. In such a scenario the only field lines that will be closed will satisfy the relation,

(8.8)\begin{equation} \frac{n}{m}=\frac{B_{\bot}}{{\rm \pi} R B_{z0}} \left(1-\frac{R}{a}\right)L,\quad n,m \ \textrm{integers,} \end{equation}

while lines not satisfying this relation will be surface filling. Nevertheless the magnetic field lines lie on cylindrical surfaces, thus one can calculate the total flux through a circular surface lying on the plane $z$ and bounded by the radius $R$. The magnetic flux in this case will simply be

(8.9)\begin{equation} \varPhi = \int \boldsymbol{B} \cdot \textrm{d} \boldsymbol{S} = {\rm \pi}R^2 B_{z0}. \end{equation}

The $\chi$ function can now be calculated according to (5.9) to yield the value

(8.10)\begin{equation} \chi = \tfrac{1}{2} B_{z0} R^2. \end{equation}

Solving (5.14) for $\eta$ we obtain the following non-unique solution:

(8.11)\begin{equation} \eta = \phi-\frac{2 B_{\bot}}{B_{z0}} \left(1-\frac{R}{a}\right) \frac{z}{R}.\end{equation}

Finally we solve (5.8) for $\mu$, here we suggest the following simple and non-unique solution:

(8.12)\begin{equation} \mu = \frac{\rho}{B_{z0}} z. \end{equation}

Thus, $\mu$ surfaces are just $z$ planes. Notice that since we have identified the planes $z=0$ and $z=L$, $\mu$ is non-single-valued. The same can be said of $\eta$ which is doubly non-single-valued in both the $z$ and $\phi$ directions.

8.2.2. The velocity field

A stationary velocity field $\boldsymbol {v}$ must satisfy the stationary versions of (2.1) and (2.3),

(8.13)\begin{gather} \boldsymbol{\nabla} \times (\boldsymbol{v} \times \boldsymbol{B}) = 0, \end{gather}

(8.14)\begin{gather}\boldsymbol{\nabla} \boldsymbol{\cdot} (\rho \boldsymbol{v} ) = 0. \end{gather}

Such a velocity field can be constructed using the labels $\mu$ and $\chi$ (see (6.19) of Yahalom & Lynden-Bell (Reference Yahalom and Lynden-Bell2008)),

(8.15)\begin{equation} \boldsymbol{v} = k \frac{\boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \chi}{\rho}, \end{equation}

where $k$ is a dimensional constant that we will choose such that $k={v_0}/{a}$. Plugging in $\mu$ from (8.12) and $\chi$ from (8.10) we arrive at the simple expression,

(8.16)\begin{equation} \boldsymbol{v} = v_0 \frac{R}{a} \hat \phi. \end{equation}

This expression can be shown to solve (2.1) and (2.3) by direct substitution. The stationary version of (2.4) is given by

(8.17)\begin{equation} \rho (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla})\boldsymbol{v} ={-}\boldsymbol{\nabla} p + \frac{(\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B}}{4 {\rm \pi}}. \end{equation}

This can be solved by the pressure function,

(8.18a,b)\begin{equation} p(R) = \rho \left[\frac{B_{\bot}^2}{\rm \pi} \left(3 \frac{R}{a} - \frac{R^2}{a^2}-\ln \left(\frac{R}{a}\right)-2\right) + \frac{1}{2} v_0^2 \left(\frac{R^2}{a^2}-1\right)\right],\quad p(a)=0. \end{equation}

8.2.3. Cross-helicities

We now can calculate the cross-helicity using (6.13), in which we assume uniform specific entropy such that $\boldsymbol {v}_t = \boldsymbol {v}$. Inserting (8.16) and (8.7) into (6.13) we arrive at the expression

(8.19)\begin{equation} \left|H_{CNB} \right| = \frac{\rm \pi}{3} v_0 B_{\bot} L a^2. \end{equation}

The ‘generalized’ magnetic fields are calculated using the label scalars as follows:

(8.20a,b)\begin{equation} \boldsymbol{B}_\chi = \boldsymbol{\nabla} \mu \times \boldsymbol{\nabla} \eta = \frac{\rho}{B_{z0}} \left[\frac{2 z B_{\bot}}{B_{z0}R^2}\hat \phi - \frac{1}{R} \hat R \right],\quad \boldsymbol{B}_\eta \equiv \boldsymbol{\nabla} \chi \times \boldsymbol{\nabla} \mu ={-} \rho R \hat \phi, \end{equation}

in which we have used $\mu$ given in (8.12), $\chi$ given in (8.10) and $\eta$ given in (8.11). This can be plugged into the generalized cross-helicities defined in (6.27a,b) to yield the results

(8.21a,b)\begin{equation} \left|H_{CNB \chi} \right| = 2 {\rm \pi}\rho v_0 \frac{B_{\bot}}{B_{z0}^2} L^2,\quad \left|H_{CNB \eta}\right| = \frac{\rm \pi}{2} \rho v_0 L a^3. \end{equation}

Now consider the case that the plasma is initially stationary with the above described density, velocity and magnetic field. Then the plasma is heated such that the density and entropy are not uniform anymore, the pressure thus changes and does not satisfy (8.18a,b). The plasma is moved from stationary equilibrium and dynamical processes occur (instabilities) in which the velocity and magnetic field are not stationary anymore. Nevertheless, if the evolution satisfies the ideal MHD equation of motion, the values for the cross-helicities given in (8.19) and (8.21a,b) are unmodified and all developing processes and instabilities must satisfy the constraints described in § 8.1.