**Appendix**

Let us calculate the divergences from each of the fluxes *J*<sup>2</sup> � *J*7, whose explicit form is represented by expressions (10a)–(10g) (*divJ*<sup>1</sup> ¼ 0). In the preconversion process we will take into account the invariance of *n=H* and the first adiabatic invariant *<sup>=</sup>mu* <sup>¼</sup> *<sup>m</sup>ν*<sup>2</sup> <sup>⊥</sup>*=*2*<sup>H</sup>* , as well as the corresponding preconversion of the divergence from the vector product *div A*½ �¼ , *B* ð Þ *B* � *rotA* ð Þ *A* � *rotB* .

For the flow divergence *J*<sup>2</sup> we obtain

$$\operatorname{div}\overleftarrow{V}\_{2} = \operatorname{div}\left(nc\frac{\left[\overleftarrow{E},\overleftarrow{H}\right]}{H^{2}}\right) = \frac{nc}{H^{2}}\operatorname{div}\left[\overleftarrow{E},\overleftarrow{H}\right] + \left[\overleftarrow{E},\overleftarrow{H}\right] \operatorname{div}\left(\frac{nc}{H^{2}}\right) = -\frac{nc}{H^{2}}\left(\overleftarrow{E}rot\overleftarrow{H}\right) - \left(\overleftarrow{E}\right)\left(\overleftarrow{E},\overleftarrow{H}\right),\tag{27}$$

$$-\frac{nc}{H}\left(\left[\overleftarrow{E},\overleftarrow{H}\right]\overleftarrow{V}\frac{1}{H}\right) = \frac{2}{m\nu\_{\perp}^{2}}\left(\overleftarrow{E}\overleftarrow{j}\_{m}\right) - \frac{nc}{H^{3}}\operatorname{\div}\left[\overleftarrow{H},\overleftarrow{V}\overleftarrow{H}\right] = \frac{2}{m\nu\_{\perp}^{2}}\left(\overleftarrow{E}\overleftarrow{j}\_{m}\right) - \frac{2}{m\nu\_{\perp}^{2}}\left(\overleftarrow{E}\overleftarrow{j}\_{\mathcal{B}^{\*}}\right). \tag{27}$$

where

$$\stackrel{\leftarrow}{j}\_{m} = -\frac{nc}{H}\mu rot\stackrel{\leftarrow}{H}, j\_{gr} = \frac{nc}{H^2}\mu \left[\stackrel{\leftarrow}{H}, \stackrel{\leftarrow}{\nabla}H\right].$$

Let's calculate the divergence from the *J*<sup>3</sup> flow:

*div J*<sup>3</sup> ¼ *div n mcν*<sup>2</sup> ⊥ <sup>2</sup>*eH*<sup>3</sup> *<sup>H</sup>* , ∇ *H* � � h i <sup>¼</sup> *<sup>n</sup> mcν*<sup>2</sup> ⊥ <sup>2</sup>*eH*<sup>3</sup> *div H* , ∇ *H* h i <sup>þ</sup> *<sup>H</sup>* , ∇ *H* h i<sup>∇</sup> *n mcν*<sup>2</sup> ⊥ <sup>2</sup>*eH*<sup>3</sup> <sup>¼</sup> ¼ � *<sup>p</sup>*<sup>⊥</sup> *eH*<sup>3</sup> <sup>∇</sup> *HrotH* � *H* , ∇ *H* h i*<sup>n</sup> mcν*<sup>2</sup> ⊥ <sup>2</sup>*eH*<sup>2</sup> <sup>∇</sup> 1 *H* ¼ ¼ *j <sup>m</sup>*∇ � � *H eH* <sup>þ</sup> *<sup>n</sup> mcν*<sup>2</sup> ⊥ <sup>2</sup>*eH*<sup>4</sup> *<sup>H</sup>* , ∇ *H* h i <sup>¼</sup> <sup>2</sup> *meν*<sup>2</sup> *<sup>F</sup> <sup>m</sup> j m* � �, (28)

The final result in formulas (27) and (28) correspond to formulas (11) and (13). where *F <sup>m</sup>* ¼ �*μ*∇ *H:*

Similarly, transform the divergences from the fluxes *J*<sup>4</sup> � *J*7, we obtain

$$\stackrel{\smile}{div}\stackrel{\smile}{J}\_4 = -\frac{2}{m\epsilon\nu\_\perp^2} \left( \stackrel{\smile}{j}\_m \stackrel{\smile}{\nabla} \frac{p\_\perp}{n} \right) + \frac{2}{m\epsilon\nu\_\perp^2} \left( \stackrel{\smile}{j}\_{\mathcal{S}^\Gamma} \stackrel{\smile}{\nabla} \frac{p\_\perp}{n} \right). \tag{29}$$

$$
\stackrel{\leftarrow}{div}\stackrel{\leftarrow}{J}\_{\mathfrak{5}} = -\frac{2}{m\epsilon\nu\_{\perp}^{2}} \left( \stackrel{\leftarrow}{F}\_{c} \stackrel{\leftarrow}{j}\_{m} \right) + \frac{2}{m\epsilon\nu\_{\parallel}^{2}} \left( \stackrel{\leftarrow}{j}\_{c} \stackrel{\leftarrow}{\nabla} \frac{p\_{\parallel l}}{n} \right), \tag{30}
$$

$$
\overrightarrow{div}\,\overleftarrow{J}\_{6} = -\frac{2}{m\epsilon\nu\_{\perp}^{2}}\left(\overleftarrow{j}\_{m}\overleftarrow{\nabla}\frac{p\_{\varPi}}{n}\right) + \frac{2}{m\epsilon\nu\_{\perp}^{2}}\left(\overleftarrow{j}\_{g^{r}}\overleftarrow{\nabla}\frac{p\_{\varPi}}{n}\right),\tag{31}
$$

$$
\stackrel{\leftarrow}{\partial} \stackrel{\leftarrow}{\partial} \stackrel{\leftarrow}{J} = -\frac{n}{H} \left( \stackrel{\leftarrow}{H} \stackrel{\leftarrow}{\nabla} \nu\_{\text{II}} \right) = \left( \stackrel{\leftarrow}{\tilde{e}\_{1}} \stackrel{\leftarrow}{\nabla} \nu\_{\text{II}} \right) \tag{32}
$$

In (29)-(32) it is necessary to transform the gradient terms from and. To do this, we use the invariance of *<sup>p</sup>*⊥*=<sup>n</sup>* � �, *pII=<sup>n</sup>* � � and *<sup>ν</sup>II*. Since

$$
\overleftarrow{\nabla} \frac{p\perp}{n} = \overleftarrow{\nabla} \frac{\overleftarrow{m} \nu\_{\perp}^{2}}{2} = \overleftarrow{\nabla} \frac{\overleftarrow{m} \nu\_{\perp}^{2}}{2H} H = \mu \overleftarrow{\nabla} H = -\overleftarrow{F}\_{m},
$$

Then

$$\left(\stackrel{\leftarrow}{j}\_{m}\stackrel{\leftarrow}{\nabla}\frac{p\_{\perp}}{n}\right) = -\left(\stackrel{\leftarrow}{j}\_{m}\stackrel{\leftarrow}{F}\_{m}\right) \tag{33}$$

*Application of Onsager and Prigozhin Variational Principles of Nonequilibrium… DOI: http://dx.doi.org/10.5772/intechopen.103116*

and

$$\left(\stackrel{\leftarrow}{j}\_{gr}\stackrel{p}{\nabla}\frac{p\_{\perp}}{n}\right) = -\left(\stackrel{\leftarrow}{j}\_{gr}\stackrel{\leftarrow}{F}\_{m}\right).\tag{34}$$

In a stationary magnetic field it is true with the accuracy of the drift approximation [21]

$$\frac{d\nu\_{\rm II}}{dt} = \frac{e}{m} \left( \stackrel{\leftarrow}{E} \stackrel{\leftarrow}{e}\_1 \right) + \frac{\nu\_\bot^2}{2} div \stackrel{\leftarrow}{e}\_1. \tag{35}$$

Assume that the first term in (35) is zero (the electric field is perpendicular to the magnetic field). Convert the second term in (35)

$$\frac{\nu\_{\perp}^{2}}{2}\text{div}\,\overleftarrow{\boldsymbol{e}}\_{1} = \frac{\nu\_{\perp}^{2}}{2}\text{div}\,\overleftarrow{\boldsymbol{H}} = \frac{\nu\_{\perp}^{2}}{2}\text{div}\,\overleftarrow{\boldsymbol{H}} - \frac{\nu\_{\perp}^{2}}{2\boldsymbol{H}^{2}}\left(\overleftarrow{\boldsymbol{H}}\overleftarrow{\boldsymbol{\nabla}}\boldsymbol{H}\right) = -\frac{\mu}{m}\left(\overleftarrow{\boldsymbol{e}}\_{1}\overleftarrow{\boldsymbol{\nabla}}\boldsymbol{H}\right) = -\frac{\mathbf{1}}{m}\left(\overleftarrow{\boldsymbol{e}}\_{1}\overleftarrow{\boldsymbol{F}}\_{m}\right).$$

Given this transformation (35) will take the form

$$\frac{d\nu\_{\rm II}}{dt} = \frac{1}{m} \left(\stackrel{\leftarrow}{e}\_1 \stackrel{\leftarrow}{F}\_m\right). \tag{36}$$

In addition, for the constant magnetic field in the drift approximation it is true

$$\frac{d\nu\_{\rm II}}{dt} = \frac{\partial\nu\_{\rm II}}{\partial t} + \left(\stackrel{\leftarrow}{U}\stackrel{\leftarrow}{\nabla}\right)\stackrel{\leftarrow}{\nabla}\nu\_{\rm II} = \left(\nu\_{\rm II}\stackrel{\leftarrow}{\mathbf{e}}\_{1} + \stackrel{\leftarrow}{U}\_{dr.}\right)\stackrel{\leftarrow}{\nabla}\nu\_{\rm II} \approx \nu\_{\rm II}\left(\stackrel{\leftarrow}{\mathbf{e}}\_{1}\stackrel{\leftarrow}{\nabla}\nu\_{\rm II}\right). \tag{37}$$

By equating (36) and (37), we obtain

$$
\stackrel{\leftarrow}{e}\_{1}\nu\_{\varPi}\stackrel{\leftarrow}{\nabla}\nu\_{\varPi} = \frac{1}{m}\left(\stackrel{\leftarrow}{e}\_{1}\stackrel{\leftarrow}{F}\_{m}\right).\tag{38}
$$

Let's write the gradient from

$$
\overleftarrow{\nabla} \left( \frac{p\_{II}}{n} \right) = \overleftarrow{\nabla} m \nu\_{II}^2 = 2m \nu\_{II} \overleftarrow{\nabla} \nu\_{II},
$$

whence, taking into account, we have

$$
\overleftarrow{j}\_{g\text{r.}} \overleftarrow{\nabla} \frac{p\_{\text{II}}}{n} = 2e \left( \overleftarrow{E} \overleftarrow{j}\_{g\text{r.}} \right) + 2 \left( \overleftarrow{F}\_{m} \overleftarrow{j}\_{g\text{r.}} \right), \tag{39}
$$

$$
\stackrel{\leftarrow}{j}\_{m.} \stackrel{\leftarrow}{\nabla} \frac{p\_{\text{II}}}{n} = 2e \left( \stackrel{\leftarrow}{E} \stackrel{\leftarrow}{j}\_{m.} \right) + 2 \left( \stackrel{\leftarrow}{F}\_{m} \stackrel{\leftarrow}{j}\_{m.} \right), \tag{40}
$$

$$
\stackrel{\leftarrow}{j}\_{c.} \stackrel{\leftarrow}{\nabla} \frac{p\_{\Pi}}{n} = 2\epsilon \left( \stackrel{\leftarrow}{E} \stackrel{\leftarrow}{j}\_{c.} \right) + 2 \left( \stackrel{\leftarrow}{F}\_{m} \stackrel{\leftarrow}{j}\_{c.} \right), \tag{41}
$$

$$\left(\overleftarrow{\mathcal{e}}\_{1}\overleftarrow{\nabla}\nu\_{\mathrm{II}}\right) = \frac{\mathcal{e}}{m\nu\_{\mathrm{II}}}\left(\overleftarrow{\mathcal{E}}\overleftarrow{\mathcal{e}}\_{1}\right) + \frac{\left(\overleftarrow{\mathcal{e}}\_{1}\overleftarrow{\mathcal{F}}\_{m}\right)}{m\nu\_{\mathrm{II}}},\tag{42}$$

By substituting the values of (33), (34), (39)-(42), into (29)-(32), we obtain the expressions (13), (14), (15) and (16) presented in Section 3, respectively.

If a time-varying electric field acts in the plasma, the crossed *E* , *H* fields produce an acceleration of electric drift

$$\frac{d\overleftarrow{V}\_E}{dt} = c \frac{\left[\overleftarrow{E}, \overleftarrow{H}\right]}{H^2} \tag{43}$$

creating an inertial force *F iner:* ¼ �*m ν : E :*

In the drift approximation, the electric field *E* and its rate of change are limited by ð Þ *cE=<sup>H</sup>* < <*<sup>V</sup>* and ð Þ *<sup>∂</sup>E=∂t*< <*E=TL* (*TL* is the period of Larmor's rotation).

The force (43) causes drift with speed

$$
\stackrel{\leftarrow}{\nu}\_P = \frac{mc^2}{H^2} \stackrel{\leftarrow}{E} \tag{44}
$$

and leads to the occurrence of electric polarization current

$$
\stackrel{\leftarrow}{j}\_P = n\epsilon\\\stackrel{\leftarrow}{\nu}\_P = \frac{nmc^2}{H^2} \stackrel{\leftarrow}{E} \tag{45}
$$

In (44) and (45) we took into account the equality to zero of the scalar product *E* � *e* 1 � �. According to (45) we have

$$\frac{\stackrel{\leftarrow}{\partial \dot{E}}}{\partial t} = \frac{H^2}{nmc^2} \stackrel{\leftarrow}{j}\_p. \tag{46}$$

Since

$$(n\_e - n\_i) = \frac{1}{4\pi e} \stackrel{\leftarrow}{div} \stackrel{\leftarrow}{E} \dots$$

then

$$\frac{\partial}{\partial t}(n\_{\epsilon} - n\_{i}) = \frac{1}{4\pi\epsilon} \text{div}\,\frac{\partial \stackrel{\leftarrow}{E}}{\partial t}.\tag{47}$$

Substituting the values of from (46) into (47), we obtain

$$\frac{\partial (n\_e - n\_i)}{\partial t} = \frac{1}{4\pi e} \operatorname{div} \left[ \frac{H^2}{nmc^2} \stackrel{\leftarrow}{j}\_P \right].$$

Calculating the divergence from the expression in square brackets of the last expression gives

$$\frac{\partial}{\partial \boldsymbol{\beta}} (\boldsymbol{n}\_{\boldsymbol{\epsilon}} - \boldsymbol{n}\_{\boldsymbol{i}}) = \frac{H^2}{4\pi emc^2} \boldsymbol{d}\boldsymbol{i}\boldsymbol{v}\boldsymbol{\hat{j}}\_{\boldsymbol{P}} - \frac{H^2}{\pi meemc^2} \frac{1}{m\nu\_{\perp}^2} \left(\boldsymbol{\hat{j}}\,\_{\boldsymbol{P}}\boldsymbol{\hat{F}}\_{\boldsymbol{m}}\right). \tag{48}$$

In the derivation of (48) the spatial derivatives of *E* were neglected with the accuracy of the drift approximation. Let us now calculate the value of *div j <sup>P</sup>*. To do this, we substitute in Maxwell's equation

*Application of Onsager and Prigozhin Variational Principles of Nonequilibrium… DOI: http://dx.doi.org/10.5772/intechopen.103116*

$$\stackrel{\rightharpoonup}{rot\overset{\rightharpoonup}{H}} = \frac{1}{c} \frac{\stackrel{\rightharpoonup}{\partial \dot{E}}}{\partial t} + \frac{4\pi}{c} \stackrel{\rightharpoonup}{j}\_P$$

value of the current *j <sup>P</sup>* from (45) and after simple transformations we obtain

$$
rot \overleftarrow{H} = \frac{\mathbf{a}}{c} \overleftarrow{E},
$$

where æ <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>4</sup>*πnmc*<sup>2</sup> *<sup>H</sup>*<sup>2</sup> [16]. From which we get

$$\frac{\partial \overleftarrow{E}}{\partial t} = \frac{c}{\mathfrak{e}} rotH. \tag{49}$$

Substituting the value of the derivative according to (49) and (45), we obtain

$$
\stackrel{\leftarrow}{j}\_P = \frac{nmc^3}{\left(H^2 + 4\pi nmc^2\right)} rot\stackrel{\leftarrow}{H}.\tag{50}
$$

Let's calculate the divergence from the right and left parts of (50), we get

$$d\dot{\upsilon}\stackrel{\leftarrow}{\dot{j}}\_P = \frac{4}{m\nu\_\perp^2} \left(\stackrel{\leftarrow}{\dot{j}}\_P \cdot \stackrel{\leftarrow}{F}\_m\right). \tag{51}$$

After substituting in (48) the value of, according to (51), we finally obtain expression (17) for the derivative of the concentration difference ð Þ *ne* � *ni* given in Section 3,

$$\frac{\partial}{\partial t}(n\_{\epsilon} - n\_{i}) = -\frac{4H^{2}}{\left(H^{2} + 4\pi mmc^{2}\right)} \frac{\left(\stackrel{\leftarrow}{j}\_{\,\,P}\stackrel{\leftarrow}{F}\_{m}\right)}{em\nu^{2}}.$$
