Generalized toroidal angular momentum

Next, we work in cylindrical coordinates $(R, \varphi, Z)$ and prove that the generalized momentum conjugating to the toroidal angle $\varphi$ is a constant of motion for axisymmetric electromagnetic field. The generalized momentum conjugating to $\varphi$ is defined by

$$P_{\varphi} = \left( \frac{\partial \mathcal{L}}{\partial \dot{\v... ..., y, \alpha, \dot{R}, \dot{Z}, \dot{v}_{\parallel}, \dot{y}, \dot{\alpha}, t} .$$ (36)

Using the Lagrangian given in Eq. (1), Eq. (36) is written as

$$P_{\varphi} = \frac{\partial}{\partial \dot{\varphi}} \left[ \left( \frac{Z e}{... ...uremath{\boldsymbol{b}} \right) \cdot \dot{\ensuremath{\boldsymbol{X}}} \right]$$

$$= \left( \frac{Z e}{c} \ensuremath{\boldsymbol{A}} + m v_{\parallel... ...\cdot \frac{\partial \dot{\ensuremath{\boldsymbol{X}}}}{\partial \dot{\varphi}}$$ (37)

Noting that $\ensuremath{\boldsymbol{X}} = \ensuremath{\boldsymbol{X}} (R, Z, \varphi)$, we obtain

$$\dot{\ensuremath{\boldsymbol{X}}} = \frac{\partial \ensuremath{\b... ... + \frac{\partial \ensuremath{\boldsymbol{X}}}{\partial \varphi} \dot{\varphi},$$ (38)

from which we get

$$\frac{\partial \dot{\ensuremath{\boldsymbol{X}}}}{\partial \dot{\varphi}} = \frac{\partial \ensuremath{\boldsymbol{X}}}{\partial \varphi} .$$ (39)

Further we note that

$$\ensuremath{\boldsymbol{X}} (R, Z, \varphi) = R \widehat{\ensurem... ...bol{X}}}{\partial \varphi} = R \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi},$$ (40)

where $\hat{e}_{\varphi}$ is the toroidal unit vector. Thus we obtain that

$$\frac{\partial \dot{\ensuremath{\boldsymbol{X}}}}{\partial \dot{\varphi}} = R \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi},$$ (41)

Using this, Eq. (37) is written as

$$P_{\varphi} = \left( \frac{Z e}{c} \ensuremath{\boldsymbol{A}} + m v_{\parallel... ...\boldsymbol{b}} \right) \cdot R \widehat{\ensuremath{\boldsymbol{e}}}_{\varphi}$$

$$= \frac{Z e}{c} A_{\varphi} R + m v_{\parallel} \frac{R B_{\varphi}}{B}$$

Define $\psi = A_{\varphi} R$, then the above equation is written as

$$P_{\varphi} = \frac{Z e}{c} \psi + m v_{\parallel} \frac{R B_{\varphi}}{B},$$ (42)

which agrees with Eq. (17) in Porcelli's paper[1]. Next we calculate the total time derivative of $P_{\varphi}$, which is given by the Euler-Lagrangian equation corresponding to $\varphi$,

$$\dot{P}_{\varphi} = \frac{\partial \mathcal{L}}{\partial \varphi} .$$ (43)

In order to calculate the partial derivative of $\mathcal{L}$ with respect to $\varphi$, we write $\ensuremath{\boldsymbol{X}}$ in terms of the cylindrical coordinate,

$$\ensuremath{\boldsymbol{X}} = R \widehat{\ensuremath{\boldsymbol{e}}}_R (\varphi) + Z \widehat{\ensuremath{\boldsymbol{e}}}_Z,$$ (44)

the time derivative of which is

$$\dot{\ensuremath{\boldsymbol{X}}} = \dot{R} \widehat{\ensuremath{... ...{\boldsymbol{e}}}_{\varphi} + \dot{Z} \widehat{\ensuremath{\boldsymbol{e}}}_Z .$$ (45)

(Note that $\partial \dot{\ensuremath{\boldsymbol{X}}} / \partial \varphi \neq 0$.) Then we have

$$\ensuremath{\boldsymbol{A}} \cdot \dot{\ensuremath{\boldsymbol{X}}} = A_R \dot{R} + A_{\varphi} R \dot{\varphi} + A_z \dot{Z},$$ (46)

from which we obtain

$$\frac{\partial ( \ensuremath{\boldsymbol{A}} \cdot \dot{\ensuremath{\boldsymbol{X}}})}{\partial \varphi} = \frac{\partial A_R}{\partial \varphi} \dot{R} + \frac{\partial A_... ...artial \varphi} R \dot{\varphi} + \frac{\partial A_z}{\partial \varphi} \dot{Z}$$

$$= \left( \frac{\partial A_R}{\partial \varphi} \widehat{\ensuremath... ...{\ensuremath{\boldsymbol{e}}}_Z \right) \cdot \dot{\ensuremath{\boldsymbol{X}}}$$ (47)

Similarly, we obtain

$$\frac{\partial ( \ensuremath{\boldsymbol{b}} \cdot \dot{\ensurema... ...{\ensuremath{\boldsymbol{e}}}_Z \right) \cdot \dot{\ensuremath{\boldsymbol{X}}}$$ (48)

Using these, the partial derivative of $\mathcal{L}$ with respect to $\varphi$ can be calculated as

$$\frac{\partial \mathcal{L}}{\partial \varphi} = \left[ \frac{Z e}... ... \left( \frac{1}{\Omega} \right) - Z e \frac{\partial \phi}{\partial \varphi} .$$ (49)

The second last term on the right-hand side of Eq. (49) can be further calculated as

$$y \dot{\alpha} \frac{\partial}{\partial \varphi} \left( \frac{1}{\Omega} \right) = y \dot{\alpha} \left( - \frac{1}{\Omega^2} \right) \frac{\partial \Omega}{\partial \varphi}$$

$$= y \dot{\alpha} \left( - \frac{1}{B \Omega} \right) \frac{\partial B}{\partial \varphi} .$$ (50)

Using $y = \mu B$ and $\dot{\alpha} = \Omega$ in the above equation, we obtain

$$y \dot{\alpha} \frac{\partial}{\partial \varphi} \left( \frac{1}{\Omega} \right) = - \mu \frac{\partial B}{\partial \varphi} .$$ (51)

Using Eq. (51) in Eq. (49) yields

$$\frac{\partial \mathcal{L}}{\partial \varphi} = \left[ \frac{Z e}... ...ac{\partial B}{\partial \varphi} - Z e \frac{\partial \phi}{\partial \varphi} .$$ (52)

For toroidal symmetrical equilibrium, the partial derivatives with respect to $\varphi$ are all zeros. In this case Eq. (52) reduces to $\partial \mathcal{L} / \partial \varphi = 0$. Using the Euler-Lagrange equation, we obtain $\dot{P}_{\varphi} = 0$, i.e., $P_{\varphi}$ is a constant of the motion in symmetrical field. [Note in passing that Eq. (52) is different from Eq. (32) in Porcelli's paper[1], which is given by

$$\frac{\partial \mathcal{L}}{\partial \varphi} = \left[ \frac{Z e}... ...ac{\partial B}{\partial \varphi} - Z e \frac{\partial \phi}{\partial \varphi} .$$ (53)

Porcelli's equation is obviously wrong since it can not recover the correct result $\dot{P}_{\phi} = 0$ for axisymmetrical electromagnetic field.]