Graduate Difficulties with derivative of a vector [Landau Textbook]

[GrimGuy]

Hi guys, I'm having trouble computing a pass 1 to 106.15. It's in the pictures.

So, what a have to do is the derivative of $f$ with respect to time and coordinates. Then I need to rearrange the terms to find the equation 106.15. I am using the following conditions. $r$ vector varies in space and the vector $r_a$ varies in time.

I did that:

This calculus is from. (Section 106)

 

[etotheipi]

The spatial dependence of $f$ is in the argument $\mathbf{r}$, whilst the time dependence is in the source positions $\mathbf{r}_a$\begin{align*}
\partial_{\alpha} |\mathbf{r} - \mathbf{r}_a| &= \frac{x_{\alpha} - x_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} = n_{a \alpha}

\end{align*}Then taking the time derivative\begin{align*}
\partial_t \partial_{\alpha} |\mathbf{r} - \mathbf{r}_a| = \partial_t \left( \frac{x_{\alpha} - x_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} \right) &= \frac{-v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} + (x_{\alpha} - x_{a \alpha}) \partial_t \left( \frac{1}{|\mathbf{r} - \mathbf{r}_a|} \right) \\

&= \frac{-v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} + \frac{-(x_{\alpha} - x_{a \alpha})}{|\mathbf{r} - \mathbf{r}_a|^2} \left( \frac{(\mathbf{r} - \mathbf{r}_a) \cdot (-\mathbf{v}_a )}{|\mathbf{r} - \mathbf{r}_a|} \right) \\

&= \frac{-v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} + \frac{n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a)}{|\mathbf{r} - \mathbf{r}_a|}

\end{align*}Hence if $f = -(k/2) \sum_a m_a |\mathbf{r} - \mathbf{r}_a |$, then\begin{align*}

\frac{\partial^2 f}{\partial t \partial x^a} &= \frac{k}{2} \sum_a m_a \left[ \frac{v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} - \frac{n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a)}{|\mathbf{r} - \mathbf{r}_a|} \right]\end{align*}and consequently\begin{align*}
h_{0 \alpha} &= \frac{8k}{2c^3} \sum_a \frac{m_a v_{a\alpha}}{|\mathbf{r} - \mathbf{r}_a|} - \frac{k}{2c^3} \sum_a \left[ \frac{m_av_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} - \frac{m_a n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a)}{|\mathbf{r} - \mathbf{r}_a|} \right] \\

&= \frac{k}{2c^3} \sum_a \frac{m_a}{|\mathbf{r} - \mathbf{r}_a|} \left[ 7v_{a \alpha} + n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a ) \right]\end{align*}

 

[GrimGuy]

The spatial dependence of $f$ is in the argument $\mathbf{r}$, whilst the time dependence is in the source positions $\mathbf{r}_a$\begin{align*}
\partial_{\alpha} |\mathbf{r} - \mathbf{r}_a| &= \frac{x_{\alpha} - x_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} = n_{a \alpha}

\end{align*}Then taking the time derivative\begin{align*}
\partial_t \partial_{\alpha} |\mathbf{r} - \mathbf{r}_a| = \partial_t \left( \frac{x_{\alpha} - x_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} \right) &= \frac{-v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} + (x_{\alpha} - x_{a \alpha}) \partial_t \left( \frac{1}{|\mathbf{r} - \mathbf{r}_a|} \right) \\

&= \frac{-v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} + \frac{-(x_{\alpha} - x_{a \alpha})}{|\mathbf{r} - \mathbf{r}_a|^2} \left( \frac{(\mathbf{r} - \mathbf{r}_a) \cdot (-\mathbf{v}_a )}{|\mathbf{r} - \mathbf{r}_a|} \right) \\

&= \frac{-v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} + \frac{n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a)}{|\mathbf{r} - \mathbf{r}_a|}

\end{align*}Hence if $f = -(k/2) \sum_a m_a |\mathbf{r} - \mathbf{r}_a |$, then\begin{align*}

\frac{\partial^2 f}{\partial t \partial x^a} &= \frac{k}{2} \sum_a m_a \left[ \frac{v_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} - \frac{n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a)}{|\mathbf{r} - \mathbf{r}_a|} \right]\end{align*}and consequently\begin{align*}
h_{0 \alpha} &= \frac{8k}{2c^3} \sum_a \frac{m_a v_{a\alpha}}{|\mathbf{r} - \mathbf{r}_a|} - \frac{k}{2c^3} \sum_a \left[ \frac{m_av_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|} - \frac{m_a n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a)}{|\mathbf{r} - \mathbf{r}_a|} \right] \\

&= \frac{k}{2c^3} \sum_a \frac{m_a}{|\mathbf{r} - \mathbf{r}_a|} \left[ 7v_{a \alpha} + n_{a \alpha} (\mathbf{n}_a \cdot \mathbf{v}_a ) \right]\end{align*}

Thx man, I'll take a time to understand and check it. Thank you a lot.

EDIT: I checked it and everything is fine and i could undestand the whole explanation, except for one little thing. What u were considering to write it '$n_{a \alpha}$'. It is the modul of the unit vector, or it is just a simplifyed notation to write $\frac{x_{\alpha} - x_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|}$ ?

 

[etotheipi]

EDIT: I checked it and everything is fine and i could undestand the whole explanation, except for one little thing. What u were considering to write it '$n_{a \alpha}$'. It is the modul of the unit vector, or it is just a simplifyed notation to write $\frac{x_{\alpha} - x_{a \alpha}}{|\mathbf{r} - \mathbf{r}_a|}$ ?

Not the modulus, it's the second option: the $\alpha$ component of the vector $\mathbf{n}_a := (\mathbf{r} - \mathbf{r}_a)/|\mathbf{r} - \mathbf{r}_a|$, as L&L define below the bit you posted. So $n_{a \alpha} := (\mathbf{r} - \mathbf{r}_a)_{\alpha}/|\mathbf{r} - \mathbf{r}_a| = (x_{\alpha} - x_{a \alpha})/|\mathbf{r} - \mathbf{r}_a|$

I'll admit it's a slightly confusing notation on the authors' part, using $a$ to enumerate the different sources and $\alpha$ to denote vector components