I was studying Landau's Classical Field Theory: Volume 2 and came across the Lagrangian for a system of charged particles, up to first-order post-Newtonian corrections ($65), which can be expressed as:
\begin{equation} \mathcal{L} = \mathcal{L_{0}} + \mathcal{L_{2}} \sum_{a} \frac{m_{a}v^{2}_{a}}{2} - \sum_{a > b}\frac{e_{a}e_{b}}{R_{ab}} + + \sum_{a} \frac{m_{a}v^{4}_{a}}{8c^{2}} + \sum_{a > b}\frac{e_{a}e_{b}}{2c^{2}R_{ab}}\left[\left(\vec{v}_{a} \cdot \hat{n}_{ab}\right)\left(\vec{v}_{b} \cdot \hat{n}_{ab}\right) \right] \, , \end{equation} where e_{a} is the charge of the particle and $r_{a}$ and $v_{a}$ are the position and velocity vector of each charge on the inertial frame, respectively. Moreover $r_{ab} = |r_{a} - r_{b}|$, $\hat{n}_{ab} = \vec{r}_{ab}/r_{ab}$
My approach to finding the Hamiltonian of this system is to use: \begin{equation} \mathcal{H} = v \frac{\partial \mathcal{L}}{\partial v} - \mathcal{L} \, \end{equation} where I would get the momentum from the Euler-Lagrange equation: $$ p = \frac{\partial \mathcal{L}}{\partial v}$$
However, Landau states that the third and fourth terms of the Lagrangian represent small corrections to $\mathcal{L}^{0}$. "On the other hand, we know from mechanics that for small changes of $\mathcal{L}$ and $\mathcal{H}$, the additions to them are equal in magnitude and opposite in sign". Basically, stating that. $$\mathcal{H} = \mathcal{H_{0}} + \mathcal{H_{2}} = \mathcal{H_{0}} - \mathcal{L_{2}}\, $$ where $$H_{0} = \sum_{a}\frac{p_{a}^{2}}{2m_{a}} + \sum_{a > b}\frac{e_{a}e_{b}}{R_{ab}}$$
and $L_{2}$ is simply converted to a function of $p_{a} and r_{a}$, by simply saying that $v_{a} = p_{a}/m_{a}$. But what about the 1PN correction to the momentum? Why is it negligible?
Why is that the case? Shouldn't I invert $p_{a}$ to obtain $v_{a}$ transform the Lagrangian, and then use the energy function to derive the Hamiltonian? Is this procedure that Landau introduces also valid for 2PN, 3PN, and higher-order corrections?
