is obtained as follows: $$\begin{aligned}
&v'_{x}(t)=v_{\perp}\cos(\theta+\Omega (t-\tau)),\quad v'_{y}(t)=v_{\perp}\sin(\theta+\Omega (t-\tau)),\quad v'_{3}=v_{3};\notag\\
&x'_{1}(t)=x_{1}+\frac{v_{\perp}}{\Omega}[\sin(\theta+\Omega (t-\tau))-\sin\theta],\notag\\
&x'_{2}(t)=x_{2}-\frac{v_{\perp}}{\Omega}[\cos(\theta+\Omega (t-\tau))-\cos\theta],\quad x'_{3}(t)=x_{3}+v_{3}(t-\tau),\end{aligned}$$ where $\Omega=\frac{qB_{0}}{m},v_{\perp}=\sqrt{v^{2}_{1}+v^{2}_{2}}.$ From now on, without loss of generality, we assume $m=q=1.$ Also through simple computation, it is easy to get $$\begin{aligned}
\begin{bmatrix}
 v'_{t1}\\
v'_{t2}\\
 v'_{t3}
 \end{bmatrix}=\begin{bmatrix}
 \cos\Omega(t-\tau)\quad-\sin\Omega(t-\tau)\quad 0\\
  \sin\Omega(t-\tau)\quad\cos\Omega(t-\tau)\quad\quad 0\\
  0\quad\quad\quad\quad\quad\quad 0\quad\quad\quad\quad\quad \quad\quad1
 \end{bmatrix}\begin{bmatrix}
 v'_{\tau 1}\\
 v'_{\tau 2}\\
v'_{\tau 3}
 \end{bmatrix}
 \triangleq \mathcal{R}(t-\tau)v'_{\tau},
 \end{aligned}$$

$$\begin{bmatrix}
 x'_{t1}\\
x'_{t2}\\
 x'_{t3}
 \end{bmatrix}=\begin{bmatrix}
 x_{1}\\
x_{2}\\
 x_{3}
 \end{bmatrix}+\begin{bmatrix}
 \frac{1}{\Omega}\sin\Omega(t-\tau)\quad \frac{1}{\Omega}\cos\Omega(t-\tau)-\frac{1}{\Omega}\quad\quad 0\\
  -\frac{1}{\Omega}\cos\Omega(t-\tau)+ \frac{1}{\Omega}\quad \frac{1}{\Omega}\sin\Omega(t-\tau)\quad\quad \quad 0\\
  0\quad\quad\quad\quad\quad\quad 0\quad\quad\quad\quad\quad \quad \quad \quad \quad (t-\tau)
 \end{bmatrix}\begin{bmatrix}
  v_{1}\\
 v_{2}\\
v_{3}
 \end{bmatrix}$$ $$\triangleq\begin{bmatrix}
 x_{1}\\
x_{2}\\
 x_{3}
 \end{bmatrix}+\mathcal{M}(t-\tau)\begin{bmatrix}
  v_{1}\\
 v_{2}\\
v_{3}
 \end{bmatrix}$$ $$\begin{aligned}
( x'_{t1},x'_{t2}, x'_{t3})=(x'_{\tau 1},x'_{\tau 2}, x'_{\tau 3})+
 \bigg(\frac{v'_{\tau 1}\sin\Omega(t-\tau)}{\Omega},\frac{v'_{\tau 2}\sin\Omega(t-\tau)}{\Omega},v'_{\tau 3}(t-\tau)\bigg).\end{aligned}$$

We introduce an equivalence relation, that is, for any different velocity $v'_{t_{1}},v'_{t_{2}},$ there exists an orthogonal matrix $\mathcal{O}(t_{1},t_{2})$ such that $v'_{t_{1}}=\mathcal{O}(t_{1},t_{2})v'_{t_{2}},$ we say $v'_{t_{1}}\sim v'_{t_{2}}.$ And all elements satisfying the above equivalence relation are denoted by $[v'_{t}],$ for short $v'_{t}.$

From the system (1.1)and the above equivalence relation, we can define the corresponding transform $$S^{0}_{t,\tau}(x',v')\triangleq\bigg(x'_{\tau}+\bigg(\frac{v'_{\tau 1}\sin\Omega(t-\tau)}{\Omega},\frac{v'_{\tau 2}\sin\Omega(t-\tau)}{\Omega},v'_{\tau 3}(t-\tau)\bigg),v'_{\tau}\bigg),$$ where $\mathcal{M}(t-\tau),\mathcal{R}(t-\tau)$ are defined in (2.5).

From the above equality (2.6), we can observe clearly the connection and difference on between Landau damping and cyclotron damping. Indeed, it can be reduced to 