$\newcommand{\abs}[1] {\left\vert #1 \right\vert}$
The Lagrangian is described in three parts: $L = L_P + L_R + L_I $ (particles + radiation + interaction). We use a Lagrangian density, so the action is $ S=\int L dt = \mathscr{\iint L \;dx\; dt} $
\begin{align*}
L_P &= \sum_\alpha {1 \over 2} m_\alpha \dot{x}_\alpha^2 \\
L_R &= {\epsilon_0 \over 2} \int{E^2 -c^2 B^2 dr}
\end{align*}
\begin{align*}
L_I &= \sum_\alpha q_\alpha \dot{r}_\alpha \cdot A(r_\alpha) - q_\alpha U(r_\alpha) \\
&= \int J(x) \cdot A(x) - \rho U(x) dx
\end{align*}
Rendering the integrals in Fourier space:
$\require{enclose}$
\begin{align*}
L_R &= \epsilon \, \enclose{horizontalstrike}{\int} \mathscr{ \abs{E}^2 - c^2 \abs{B}^2 } dk\\
L_I &= \enclose{horizontalstrike}\int \mathscr{j^* \cdot A + j \cdot A^*- \rho^* U- \rho U^*} dk
\end{align*}
But since $\mathscr{E =-i k U- \dot A}$ and $\mathscr{ B = i k \times A}$ we can write
$$ \mathscr{ \abs{E}^2 = k^2 \abs{U}^2 + \abs{\dot A}^2 + i U k \cdot \dot A^*} + \text{c.c.}$$
Notice that $\dot U$ does not appear in $L$
$$ 0 = \mathscr{ \frac{\partial L}{\partial U^*} = \epsilon_0 (k^2 U - i k \cdot \dot{A}) - \rho} $$
$$ \mathscr{ U = \frac{\rho}{\epsilon_0 k^2} + {i \over k} \dot{A_{||}}} $$
Use this to eliminate U:
\begin{equation*}
\mathscr{E = -{ i \rho \underline{k} \over \epsilon_0 k^2} + \dot {A_{||}} - \dot{A} = -{i \rho \underline{k} \over \epsilon_0 k^2} - \dot{A_{\perp}}}
\end{equation*}
Which lets us write
\begin{align*}
L_R &= \mathscr{ \epsilon_0 \enclose{horizontalstrike}\int {\abs{\rho}^2 \over k^2 \epsilon_0^2} + \abs{\dot{A_{\perp}}}^2 -c^2 k^2 \abs{A_{\perp}}^2 dk} \\
L_I &= \mathscr{ \enclose{horizontalstrike}\int j_{||} A_{||}^* + \text{c.c.} - {2 \abs{\rho}^2 \over \epsilon_0 k^2} -\frac{i \rho^* \dot{A_{||}}}{k} + \text{c.c.} \; dk}
\end{align*}
Observe that $ \mathscr{A}_{||}$ only appears in $\mathscr{L}_I$
\begin{align*}
\mathscr{\frac{\partial L} {\partial A_{||}^*} } &= j_{||} \\
\mathscr{\frac{\partial L}{ \partial \dot{A_{||}^*}}} &= \frac{i \rho}{k} \\
j_{||} &= {i \dot{\rho} \over k}
\end{align*}
Use this to eliminate $j_{||}$
\begin{align*}
L_I &= \mathscr{ \enclose{horizontalstrike}\int i {\dot{\rho} \over k} A_{||}^* + \text{c.c.} - 2 {\abs{\rho}^2 \over \epsilon_0 k^2} - i {\rho^* \dot{A}_{||} \over k} + \text{c.c.} + j_{\perp} \cdot A_{\perp}^* + \text{c.c.} \; dk }\\
&= \mathscr{ \enclose{horizontalstrike}\int -2 {\abs{\rho}^2 \over \epsilon_0 k^2} + {i \over k} {d \over dt} (\rho A_{||}^* - \text{c.c.}) + j_{\perp} \cdot A_{\perp}^* + \text{c.c.} \; dk }
\end{align*}
\begin{align*}
L = \sum_\alpha \frac12 m_\alpha \abs {\dot{r}_{\alpha}}^2 &- \enclose{horizontalstrike}\int {\abs{\rho}^2 \over \epsilon_0 k^2} dk \\
&+ \mathscr{ \epsilon_0 \enclose{horizontalstrike}\int \abs{\dot{A_{\perp}}}^2 - c^2 k^2 \abs{A_{\perp}}^2
dk }\\
&+ \mathscr{ \enclose{horizontalstrike} \int j_{\perp} \cdot A_{\perp}^* + \text{c.c.} \; dk}\\
&+ \mathscr{ {i \over k} {d \over dt} \enclose{horizontalstrike}\int \rho A_{||}^* -\text{c.c.} \;dk }
\end{align*}
Lagrangian dynamics is insensitive to a total time derivative, so the last term gives some freedom in choice of $A_{||}$ ("gauge").
