Lagrangian of system of bodies in PN approximation [Landau Textbook]

[GrimGuy]

Hey guy,
I'm having problems to understand the final part of this section. The book says we have the lagrangian from one particle (106.16), then we have some explanation and then the total lagrangian is given(106.17). For me is everything fine until the 106.16, then i couldn't get what is going on. What I've tried to do is, I've substituted the $h_{00}$, $h_{0\alpha}$, $h_{\alpha \beta }$ and $\phi$, $\phi_{a}$ into 106.16 and tried to find 106.17 (no sucsses). Any enlightenment on this will be extremely appreciated.

 

[etotheipi]

This does look slightly yucky! I reckon what you have to do to prove it for yourself is, like they suggest, show that $\left( \partial L_a / \partial \mathbf{r}\right)$ evaluated at $\mathbf{r} = \mathbf{r}_a$ is equal to $\partial L / \partial \mathbf{r}_a$. But I can't see a particularly nice way of doing it.

First you're going to want to work out all of $\partial h_{00} / \partial \mathbf{r}$, then $\partial h_{0\alpha} / \partial \mathbf{r}$, then $\partial h_{\alpha \beta} / \partial \mathbf{r}$. For instance, let's have a look at $\partial h_{00} / \partial \mathbf{r}$ first. We know that\begin{align*}

\partial_{\mathbf{r}} \phi &= \sum_b \frac{km_b(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3}\end{align*}Hence\begin{align*}
\partial_{\mathbf{r}} h_{00} &= \frac{2}{c^2} \partial_{\mathbf{r}} \phi + \frac{4\phi}{c^2} \partial_{\mathbf{r}} \phi + \frac{2k}{c^4} \sum_b m_b \phi_b' \frac{m_b(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3} + \frac{3k}{c^4} \sum_b m_b v_b^2 \frac{(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3}
\end{align*}which is the same as \begin{align*}

\partial_{\mathbf{r}} h_{00} = &\frac{2}{c^2} \sum_b \frac{km_b(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3} + \frac{-4k^2}{c^2} \sum_b \sum_c \frac{m_b m_c (\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3|\mathbf{r} - \mathbf{r}_c|} \\

&+ \frac{-2k^2}{c^4} \sum_b \sum_c' \frac{m_b m_c (\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3 |\mathbf{r}_b - \mathbf{r}_c|} + \frac{3k}{c^4} \sum_b \frac{m_b v_b^2 (\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3}

\end{align*}and et. cetera, assuming there's no major mistakes in the above. I really don't know if there's any simpler way. Good luck!

 

[GrimGuy]

This does look slightly yucky! I reckon what you have to do to prove it for yourself is, like they suggest, show that $\left( \partial L_a / \partial \mathbf{r}\right)$ evaluated at $\mathbf{r} = \mathbf{r}_a$ is equal to $\partial L / \partial \mathbf{r}_a$. But I can't see a particularly nice way of doing it.

First you're going to want to work out all of $\partial h_{00} / \partial \mathbf{r}$, then $\partial h_{0\alpha} / \partial \mathbf{r}$, then $\partial h_{\alpha \beta} / \partial \mathbf{r}$. For instance, let's have a look at $\partial h_{00} / \partial \mathbf{r}$ first. We know that\begin{align*}

\partial_{\mathbf{r}} \phi &= \sum_b \frac{km_b(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3}\end{align*}Hence\begin{align*}
\partial_{\mathbf{r}} h_{00} &= \frac{2}{c^2} \partial_{\mathbf{r}} \phi + \frac{4\phi}{c^2} \partial_{\mathbf{r}} \phi + \frac{2k}{c^4} \sum_b m_b \phi_b' \frac{m_b(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3} + \frac{3k}{c^4} \sum_b m_b v_b^2 \frac{(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3}
\end{align*}which is the same as \begin{align*}

\partial_{\mathbf{r}} h_{00} = &\frac{2}{c^2} \sum_b \frac{km_b(\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3} + \frac{-4k^2}{c^2} \sum_b \sum_c \frac{m_b m_c (\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3|\mathbf{r} - \mathbf{r}_c|} \\

&+ \frac{-2k^2}{c^4} \sum_b \sum_c' \frac{m_b m_c (\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3 |\mathbf{r}_b - \mathbf{r}_c|} + \frac{3k}{c^4} \sum_b \frac{m_b v_b^2 (\mathbf{r} - \mathbf{r}_b)}{|\mathbf{r} - \mathbf{r}_b|^3}

\end{align*}and et. cetera, assuming there's no major mistakes in the above. I really don't know if there's any simpler way. Good luck!

I got your idea, i'll try it. But, do you have other idea to start from 106.16 and arrive in the 106.17, this is my main go.
Thanks man.