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

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The discussion revolves around understanding the transition from the Lagrangian of a single particle to the total Lagrangian as presented in a specific section of a physics text. The initial confusion arises after the equation 106.16, where the user struggles to derive equation 106.17 by substituting various variables such as h_{00}, h_{0α}, h_{αβ}, φ, and φ_a. A suggested approach includes calculating the partial derivatives of these variables with respect to the position vector r, specifically starting with h_{00}. The user is encouraged to show that the derivative of the Lagrangian with respect to r at a specific point equals the total Lagrangian's derivative at that point. The complexity of the calculations is acknowledged, with a focus on ensuring accuracy in the derivatives to facilitate the transition from 106.16 to 106.17.
GrimGuy
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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.

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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! 😜
 
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etotheipi said:
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.
 
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