tmp (do not read this)

On the other hand we have

$$\dot{\varepsilon} = \frac{d H}{d t},$$

which $H$ is the Hamiltonian. We note that it is the energy expressed in terms of generalized coordinate and momentum that can be called Hamilton. Using the Hamilton's equation, it can be easily proved that

$$\frac{d H}{d t} = \frac{\partial H}{\partial t} .$$

Further we note that

$$\frac{\partial H}{\partial t} + \frac{\partial \mathcal{L}}{\partial t} = \frac{\partial}{\partial t} \sum_i p_i \dot{q}_i .$$ (189)

In usual case, $p_i$ and $\dot{q}_i$ are not an explicit function of time, thus we have

$$\frac{\partial H}{\partial t} + \frac{\partial \mathcal{L}}{\partial t} = 0.$$ (190)

Using these results, we have

$$\dot{\varepsilon} = - \frac{\partial L}{\partial t} .$$ (191)

Then using Eq. (35) we obtain

$$- \frac{\partial L}{\partial t} = - \left( \frac{Z e}{c} \frac{\p... ...}} + \mu \frac{\partial B}{\partial t} + Z e \frac{\partial \phi}{\partial t} .$$ (192)

$$\mathcal{L} = \left( \frac{Z e}{c} \ensuremath{\boldsymbol{A}} + ... ...frac{1}{\Omega} y \dot{\alpha} - \frac{1}{2} m v_{\parallel}^2 - y - Z e \phi .$$ (193)

$$\frac{\partial L}{\partial t} = \left( \frac{Z e}{c} \frac{\parti... ...ial \Omega}{\partial t} y \dot{\alpha} - Z e \frac{\partial \phi}{\partial t}$$

$$\frac{\partial L}{\partial t} = \left( \frac{Z e}{c} \frac{\parti... ...\partial B}{\partial t} y \dot{\alpha} - Z e \frac{\partial \phi}{\partial t}$$

$$\frac{\partial L}{\partial t} = \left( \frac{Z e}{c} \frac{\parti... ...artial B}{\partial t} \mu \dot{\alpha} - Z e \frac{\partial \phi}{\partial t}$$

The first order equation of Eq. (192) is (noting that $\ensuremath{\boldsymbol{A}}^{(0)}$, $\ensuremath{\boldsymbol{b}}^{(0)}$, and $B^{(0)}$ is time independent, thus $\partial \ensuremath{\boldsymbol{A}}^{(0)} / \partial t$, $\partial \ensuremath{\boldsymbol{b}}^{(0)} / \partial t$, and $\partial B^{(0)} / \partial t$ are all zeros)

$$- \left( \frac{\partial L}{\partial t} \right)^{(1)} = - \left( \... ...ac{\partial B^{(1)}}{\partial t} + Z e \frac{\partial \phi^{(1)}}{\partial t} .$$ (194)

In writing the above expression, we have used the fact that $\partial / \partial \varphi$ here is taken by holding constant $( \dot{\ensuremath{\boldsymbol{X}}}, \dot{v_{\parallel}}, \dot{y}, \dot{\alpha} ; \psi, \theta, v_{\parallel}, y, \alpha)$, instead of holding constantnt $( \dot{\psi}, \dot{\theta}, \dot{\varphi}, \dot{v_{\parallel}}, \dot{y}, \dot{\alpha} ; \psi, \theta, v_{\parallel}, y, \alpha)$. In this case obviously $\partial \dot{\ensuremath{\boldsymbol{R}}} / \partial \varphi = 0$. If we calculate in the second case, then we would have $\partial \dot{\ensuremath{\boldsymbol{R}}} / \partial \psi \neq 0$, since

$$\dot{\ensuremath{\boldsymbol{X}}} = \frac{\partial \ensuremath{\b... ... + \frac{\partial \ensuremath{\boldsymbol{X}}}{\partial \varphi} \dot{\varphi},$$ (195)

in which the terms such as $\partial \ensuremath{\boldsymbol{X}} / \partial \psi$ would explicitly contain $\varphi$. The second term on the right-hand side of Eq. () can be further calculated as

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

$$= y \dot{\alpha} \left( - \frac{1}{B \Omega} \right) \frac{\partial B}{\partial \psi} .$$ (196)

Then we can use $y = \mu B$ and $\dot{\alpha} = \Omega$ in the above equation, yielding

$$y \dot{\alpha} \frac{\partial}{\partial \varphi} \left( \frac{1}{\Omega} \right) = - \mu \frac{\partial B}{\partial \psi} .$$ (197)

Here I have some important comments. First, we note that $\dot{\alpha} = \Omega$ is one of the components of the Euler-Lagrange equation, thus of course can not be substituted into the original Lagrangian $\mathcal{L}$ (if we do this, we can no longer use the resulting Lagrangian as a correct Lagrangian to obtain correct Euler-Lagrange equation). In contrast to this, it is obvious we can use one component of the Euler-Lagrangian equation in another component equation. Thus we can substitute $\dot{\alpha} = \Omega$ into Eq. (196) to get Eq. (197). Second, we also substitute $y = \mu B$ into Eq. (196) in obtaining Eq. (197). This is trivial since what we do is only to rewrite the final result in a different form. However this kind of rewriting may be misleading to someone (including me) because the new form can be viewed as being written in terms of a new variable $\mu$, instead of the original variable $y$. Of course, for this case, no matter which variable the right-hand side of Eq. (197) is understood to be written in terms of, the results are both correct. But it is crucial to understand correctly which variables Lagrangian $\mathcal{L}$ is written in terms of, since different choice of variables will give different forms of perturbed Lagrangian because the perturbed Lagrangian is obtained by keeping the independent variables constant.

Now I calculate the perturbed Lagrangian. The full Lagrangian is given by Eq. (1), i.e.,

$$\mathcal{L} = \left( \frac{Z e}{c} \ensuremath{\boldsymbol{A}} + ... ...frac{1}{\Omega} y \dot{\alpha} - \frac{1}{2} m v_{\parallel}^2 - y - Z e \phi .$$

Then the perturbed and linearized version is (note that only the electromagnetic field is perturbed, the independent variables are keep constant)

$$\mathcal{L}^{(1)} = \left( \frac{Z e}{c} \ensuremath{\boldsymbol{... ...{X}}} + \left( \frac{1}{\Omega} \right)^{(1)} y \dot{\alpha} - Z e \phi^{(1)} .$$ (198)

Using

$$\left( \frac{1}{B} \right)^{(1)} = - \frac{1}{B_0^2} B^{(1)},$$ (199)

in Eq. (198), we obtain

$$\mathcal{L}^{(1)} = \left( \frac{Z e}{c} \ensuremath{\boldsymbol{... ...} - \frac{m c}{Z e} \frac{1}{B_0^2} B^{(1)} y \dot{\alpha} - Z e \phi^{(1)} .$$

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

$$\mathcal{L}^{(1)} = \left( \frac{Z e}{c} \ensuremath{\boldsymbol{... ...right) \cdot \dot{\ensuremath{\boldsymbol{X}}} - \mu B^{(1)} - Z e \phi^{(1)} .$$ (200)

My question is whether it is valid to substitute one of the Euler-Lagrangian equation $\dot{\alpha} = \Omega$ into the perturbed Lagrangian.**wrong!!**