I have the Lagrangian $$\alpha(\boldsymbol{\dot{r}} -\boldsymbol{v}(\boldsymbol{r}))^{2} + \beta \nabla \cdot \boldsymbol{v}(\boldsymbol{r}),\tag{1}$$ where $\boldsymbol{r}$ is the position and $\boldsymbol{\dot{r}}$ is velocity of a particle in a vector field $\boldsymbol{v}(\boldsymbol{r})$. To derive the equation of motion, the Euler langrange equation is applied directly to give, I think: $$2\alpha \ddot{r}_{p} = -2\alpha\dot{r}_{i}\partial_{p}v_{i}+ 2\alpha v_{i}\partial_{p} v_{i} + \beta \partial_{p} (\nabla\cdot \boldsymbol{v}).\tag{2}$$

Firstly, is this equation correct?

Secondly, when I Taylor expand $v(r)$ around $0$ for this equation, I get a different answer as to when I taylor expand $v(r)$ in the lagrangian and calculate the equation of motion from this (where they are both Taylor expanded to have to same order in $r$). Is this normal? Which one should be used for further calculations?

  • $\begingroup$ If you are getting a different answer Taylor expanding before/after, then you are either miscalculating the equation of motion, or not going sufficiently far enough out in order. Remember that when you take derivatives in deriving the EOM, you will reduce the order of the expansion. $\endgroup$ – Danny Aug 15 at 16:02
  • $\begingroup$ $p$ here is the indice, so for a three dimensional problem $p = 1, 2, 3$. $i$ is also an indice which is summed over here. $\partial_{p} \equiv \frac{\partial }{\partial r_{p}}$ $\endgroup$ – Hello Aug 16 at 8:26
  • $\begingroup$ Comment to the question (v2): There is a $-2\alpha\dot{r}_{i}\partial_{i}v_{k}$ term missing on the LHS of eq. (2). $\endgroup$ – Qmechanic Aug 22 at 8:22
  • $\begingroup$ Thank you very much for your reply! My apologies, I am still not quite seeing it, is it coming from $\frac{d}{dt}(\frac{\partial}{\partial r_{p}})$? I thought $\alpha(\dot{\boldsymbol{r}} - \boldsymbol{v}(\boldsymbol{r}))^{2} =\alpha(\dot{r}_{1}-v_{1}(\boldsymbol{r}))^{2}+\alpha(\dot{r}_{2}-v_{2}(\boldsymbol{r}))^{2}+\alpha(\dot{r}_{3}-v_{3}(\boldsymbol{r}))^{2}$ (ie: dot product of two vectors), so that for eg:$ \frac{d}{dt}(\frac{\partial L}{\partial r_{1}}) = \frac{d}{dt}(2\alpha(\dot{r}_{1} - v_{1}(\boldsymbol{r})) = 2\alpha\ddot{r_{1}}$? $\endgroup$ – Hello Aug 22 at 9:30
  • $\begingroup$ and then also $\frac{\partial L}{\partial r_{1}} = 2\alpha(\dot{r}_{1} - v_{1}(\boldsymbol{r}))(-\frac{\partial v_{1}(\boldsymbol{r})}{\partial r_{1}}) + 2\alpha(\dot{r}_{2} - v_{2}(\boldsymbol{r}))(-\frac{\partial v_{2}(\boldsymbol{r})}{\partial r_{1}}) + 2\alpha(\dot{r}_{3} - v_{3}(\boldsymbol{r}))(-\frac{\partial v_{1}(\boldsymbol{r})}{\partial r_{1}}) $ $+ \beta \frac{\partial (\nabla .\boldsymbol{v})}{\partial r_{1}}$. Where does the $\dot{r}_{i}\partial_{i}$ come from? which part of the E-L equation am I forgetting the term? $\endgroup$ – Hello Aug 22 at 9:35

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