Constants of motion in the theory of energetic particles propagating through magnetic turbulence

Abstract

To understand the motion of particles in a turbulent magnetic field is a fundamental problem in modern physics. However, it is difficult to explore this problem analytically due to the complicated structure of such fields. In the current paper we derive constants of motion based on the Lagrangian. The special cases of slab and two-dimensional turbulence are considered. The obtained results can be used to simplify equations of motions which can lead to improved non-linear theories for particle transport. The findings are also compared with previously obtained results.

Appendices

Appendix A: The Lagrangian in classical notation

Alternatively, compared to the Minkowski coordinates used in Sect. 2 of this paper, one can use classical notation. In this case the Lagrangian of a particle in the electromagnetic field has the form

$$ \mathcal{L} = - m c^{2} \sqrt{1 - v^{2} / c^{2}} + \frac{q}{c} \vec{v} \cdot \vec{A} - q \Phi $$

(106)

which can also be written as

$$\begin{aligned} \mathcal{L} &= - m c^{2} \sqrt{1 - v^{2} / c^{2}} \\ &\quad+ \frac{q}{c} \sum _{j} v_{j} A_{j} - q \Phi \quad \quad \mbox{with} \quad \quad j=1,2,3. \end{aligned}$$

(107)

The Euler-Lagrange equations now have the form

$$ \frac{d}{d t} \frac{\partial \mathcal{L}}{\partial v_{i}} = \frac{\partial \mathcal{L}}{\partial x_{i}} \quad \quad \mbox{with} \quad \quad i=1,2,3. $$

(108)

The conservation relation, previously given by Eq. (7), turns into

$$ \frac{\partial \mathcal{L}}{\partial x_{i}} = 0 \quad \quad \Rightarrow \quad \quad P_{i} = const $$

(109)

where the canonical momentum is now obtained via

$$ P_{i} = \frac{\partial \mathcal{L}}{\partial v_{i}}. $$

(110)

For the electromagnetic case (see Eq. (107)) this becomes

$$\begin{aligned} P_{i} = & \frac{\partial}{\partial v_{i}} \left [ - m c^{2} \sqrt{1 - v^{2} / c^{2}} + \frac{q}{c} \sum _{j} v_{j} A_{j} - q \Phi \right ] \\ = & \frac{1}{2} \gamma m \frac{\partial}{\partial v_{i}} \sum _{j} v_{j}^{2} + \frac{q}{c} \sum _{j} \frac{\partial v_{j}}{\partial v_{i}} A_{j} \end{aligned}$$

(111)

where we have used again the Lorentz factor $\gamma$ as given by Eq. (29). Therefore, we find for the canonical momentum

$$ P_{i} = \gamma m v_{i} + \frac{q}{c} A_{i}. $$

(112)

For the Lagrangian given by Eq. (107) we now deduce the spatial derivative needed in Eq. (108)

$$ \frac{\partial \mathcal{L}}{\partial x_{i}} = \frac{q}{c} \sum _{j} v_{j} \frac{\partial A_{j}}{\partial x_{i}} - q \frac{\partial \Phi}{\partial x_{i}}. $$

(113)

Using this and Eq. (112) in Eq. (108) yields

$$ \frac{d}{d t} \left [ \gamma m v_{i} + \frac{q}{c} A_{i} \right ] = \frac{q}{c} \sum _{j} v_{j} \frac{\partial A_{j}}{\partial x_{i}} - q \frac{\partial \Phi}{\partial x_{i}}. $$

(114)

With

$$ \frac{d A_{i}}{d t} = \frac{\partial A_{i}}{\partial t} + \sum _{j} \frac{\partial A_{i}}{\partial x_{j}} v_{j} $$

(115)

this can be written as

$$ \frac{d}{d t} \big( \gamma m v_{i} \big) = - \frac{q}{c} \frac{\partial A_{i}}{\partial t} - q \frac{\partial \Phi}{\partial x_{i}} + \frac{q}{c} \sum _{j} v_{j} \left ( \frac{\partial A_{j}}{\partial x_{i}} - \frac{\partial A_{i}}{\partial x_{j}} \right ). $$

(116)

The left-hand-side of this equation corresponds to the time-derivative of the relativistic momentum. In order to see that we found the correct equation of motion, we need to express the right-hand-side by the components of the electric and magnetic fields. To do this we consider

$$ \left ( \vec{v} \times \vec{B} \right )_{n} = \sum _{i,j} v_{i} B_{j} \epsilon _{nij} $$

(117)

as well as

$$ B_{j} = \left ( \vec{\nabla} \times \vec{A} \right )_{j} = \sum _{k, \ell} \partial _{k} A_{\ell} \epsilon _{jk\ell}. $$

(118)

Combining the latter two equations yields

$$ \left ( \vec{v} \times \vec{B} \right )_{n} = \sum _{i,j,k,\ell} \epsilon _{nij} \epsilon _{jk\ell} v_{i} \partial _{k} A_{\ell}. $$

(119)

Using the property $\epsilon _{nij} = - \epsilon _{nji} = \epsilon _{jni}$ of the Levi-Civita symbol allows us to write this as

$$ \left ( \vec{v} \times \vec{B} \right )_{n} = \sum _{i,k,\ell} \sum _{j} \epsilon _{jni} \epsilon _{jk\ell} v_{i} \partial _{k} A_{\ell}. $$

(120)

This can be simplified by employing the well-known relation (see, e.g., Boas (2006))

$$ \sum _{j} \epsilon _{jni} \epsilon _{jk\ell} = \delta _{nk} \delta _{i \ell} - \delta _{n\ell} \delta _{ik}. $$

(121)

Therewith, we find

$$\begin{aligned} \left ( \vec{v} \times \vec{B} \right )_{n} = & \sum _{i,k,\ell} \delta _{nk} \delta _{i\ell} v_{i} \partial _{k} A_{\ell} - \sum _{i,k, \ell} \delta _{n\ell} \delta _{ik} v_{i} \partial _{k} A_{\ell} \\ = & \sum _{i} \left ( v_{i} \partial _{n} A_{i} - v_{i} \partial _{i} A_{n} \right ). \end{aligned}$$

(122)

In the latter formula we rename $n \rightarrow i$ and $i \rightarrow j$ so that we obtain

$$ \big( \vec{v} \times \vec{B} \big)_{i} = \sum _{j} v_{j} \left ( \partial _{i} A_{j} - \partial _{j} A_{i} \right ). $$

(123)

This can be used in Eq. (116) to derive

$$ \frac{d}{d t} \big( \gamma m v_{i} \big) = - \frac{q}{c} \frac{\partial A_{i}}{\partial t} - q \frac{\partial \Phi}{\partial x_{i}} + \frac{q}{c} \big( \vec{v} \times \vec{B} \big)_{i}. $$

(124)

Finally we can replace the remaining potentials by the electric field as given by Eq. (22) so that

$$ \frac{d}{d t} \big( \gamma m v_{i} \big) = q \Big( \vec{E} + \frac{\vec{v}}{c} \times \vec{B} \Big)_{i} $$

(125)

in perfect agreement with the Newton-Lorentz equation.

As an example we consider the magnetostatic and two-dimensional case where $A_{i} = A_{i} (x,y)$ and, thus, it follows from Eq. (109) that

$$ \frac{\partial \mathcal{L}}{\partial z} = 0 \quad \quad \Rightarrow \quad \quad P_{z} = const. $$

(126)

Therefore, we find by using Eq. (112)

$$ P_{z} = \gamma m v_{z} + \frac{q}{c} A_{z} = const $$

(127)

in agreement with Eq. (48).

Appendix B: Alternative investigation of particles in slab turbulence

In the main part of this paper it is stated that no constant of motion can be found for particles moving through slab turbulence with mean magnetic field unless one uses guiding center coordinates (see Sects. 4.2 and 5). The considerations presented in Sect. 4 are based on Eq. (40) for the vector potential of the mean field. However, there are alternative forms of the vector potential leading to a constant mean field pointing into the $z$-direction.¹ One of such alternatives is $\vec{A} = B_{0} x \hat{y}$. For this form the total vector potential of slab turbulence and a mean field is given by

$$ \vec{A} \big( x, z \big) = \delta A_{x} \big( z \big) \hat{x} + \left [ \delta A_{y} \big( z \big) + B_{0} x \right ] \hat{y} $$

(128)

instead of Eq. (50). Therefore, there is no $y$-dependence of the total vector potential and conservation law (18) turns into

$$ \partial _{2} A^{\beta} = 0 \quad \quad \Rightarrow \quad \quad P_{2} = const. $$

(129)

Using this result in Eq. (17) yields

$$ P_{2} = m u_{2} + \frac{q}{c} A_{2} = const. $$

(130)

With the above vector potential this can be written as

$$ m u_{2} + \frac{q}{c} \left [ \delta A_{y} \big( z \big) + B_{0} x \right ] = const. $$

(131)

Alternatively, this can be written as

$$ m \gamma v_{y} + \frac{q}{c} B_{0} x + \frac{q}{c} \delta A_{y} \big( z \big) = const. $$

(132)

Or, with the help of Eq. (57), we can write this as

$$ v_{y} + \Omega x + \frac{\Omega}{B_{0}} \delta A_{y} \big( z \big) = const. $$

(133)

Although this corresponds indeed to a constant of motion, the result is not particularly useful. However, one can use again guiding center coordinates (we use the first line of Eq. (60)) to derive

$$ X + \frac{1}{B_{0}} \delta A_{y} \big( z \big) = const $$

(134)

in perfect agreement with Eq. (70). Therefore, we simply found an alternative derivation for an already known result.

Furthermore, one could use $\vec{A} = - B_{0} y \hat{x}$ for the potential of the mean magnetic field. After performing the same steps as above, one then arrives at

$$ m \gamma v_{x} - \frac{q}{c} B_{0} y + \frac{q}{c} \delta A_{x} \big( z \big) = const. $$

(135)

Using again guiding center coordinates yields Eq. (71).