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On page 4 of Landau & Lifshitz's Mechanics they say

$$L\left({v^\prime}^2\right) = L\left(v^2 + 2\bf{v \cdot} \bf{\epsilon} + \epsilon^2\right).$$ Expanding this expression in powers of $\epsilon$ and neglecting terms above the first order, we obtain

$$L\left({v^\prime}^2\right) = L\left(v^2\right) + \frac{\partial L}{\partial v^2}2~\bf{v \cdot} \epsilon\,.$$

and then on the next line:

The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $\bf v$

What is the meaning of this statement? This being a total time derivative would say $ \frac{\partial L}{\partial v^2}2\bf{v \cdot} \epsilon = C(t)$. Where do we get a linear function of $\bf v$ from this?

EDIT: The first part can be found at Expansion of a function Landau's derivation of a free particle's kinetic energy- expansion of a function? , & Trouble with Landau & Lifshitz's expansion of the Lagrangian with respect to $\epsilon$ and $v$

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    $\begingroup$ Possible duplicate of Deriving the Lagrangian for a free particle $\endgroup$ – user36790 Oct 26 '16 at 4:08
  • $\begingroup$ Observe that for Galilean invariance, it should be only dependent on constant velocity. The second term already contained $\mathbf v\,.$ Thus $\frac{\partial \mathcal L}{\partial \dot{r}^2}= \textrm{constant}\,.$ $\endgroup$ – user36790 Oct 26 '16 at 4:15
  • $\begingroup$ Yup, it's not your fault. Landau/Lifschitz really does contain a bunch of sneaky and confusing tricks, all there to make the book about 10% shorter. $\endgroup$ – knzhou Oct 26 '16 at 5:55
  • $\begingroup$ @knzhou - but Landau probably thought it was long-winded as is... $\endgroup$ – Jon Custer Oct 26 '16 at 23:32
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This is just the first term for a Taylor expansion at $v_2$ for small $2\bf{v \cdot} \bf{\epsilon} + \epsilon^2$ of the function $$L\left({v^\prime}^2\right) = L\left(v^2 + 2\bf{v \cdot} \bf{\epsilon} + \epsilon^2\right)≈ L\left(v^2\right) + \frac{\partial L}{\partial v^2}2~\bf{v \cdot} \epsilon$$ where only the lowest order (linear) terms in $\epsilon$ are retained. This means that in the first order Taylor term $\epsilon^2$ was neglected compared to $2\bf{v \cdot}\bf{\epsilon}$. With regard to the second part of your question $\mathbf v=d \mathbf r/dt$. Therefore total time derivative.

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Your first question are already answered in the links of your edit. Your second question have the following answer:

A total time derivative means that you can find a function $G$ such that: $$ dG=\sum_i \frac {\partial G}{\partial \theta _i } d\theta _i $$ Where $\theta _i$ could be every thing that depends on time: position, velocity, acceleration, etc. You want to show under what conditions: $$ dG=\frac{\partial L}{\partial v^2}\vec{v}\cdot\vec{\varepsilon} dt $$ For some funtion $G$. The right-handed side says already that $G$ cannot have any explicity dependence on $v$ or higher derivatives in $\vec {r}$, otherwise you would have a acceleration in the right-handed side because: $$ \frac{\partial G}{\partial v_i}\neq 0 \implies dG=...+\frac{\partial G}{\partial v_i}dv_i+... $$ $$ \frac{\partial G}{\partial v_i}\frac {dv_i}{dt}=\frac{\partial G}{\partial v_i} a_i $$ Then, $G$ only depends explicitly on $\vec{r}$ and $t$. Now, noting that $d\vec{r}= \vec{v} dt$ we have: $$ dG=\frac{\partial L}{\partial v^2}\vec{\varepsilon}\cdot d\vec{r} = \nabla G \cdot d \vec{r} $$ Because $G$ do not depend explicitly on $\vec{v}$ , $\nabla G$ don't depend on $\vec{v}$ as well. This implies that $\frac{\partial L}{\partial v^2}$ is a constant. Finished.

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