# Expansion in $\epsilon$ and $v^2$ dependence of the Lagrangian - Landau & Lifshitz's Mechanics [duplicate]

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?

• Possible duplicate of Deriving the Lagrangian for a free particle – user36790 Oct 26 '16 at 4:08
• Observe that for Galilean invariance, it should be only dependent on constant velocity. The second term already contained $\mathbf v\,.$ Thus $\frac{\partial \mathcal L}{\partial \dot{r}^2}= \textrm{constant}\,.$ – user36790 Oct 26 '16 at 4:15
• 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. – knzhou Oct 26 '16 at 5:55
• @knzhou - but Landau probably thought it was long-winded as is... – Jon Custer Oct 26 '16 at 23:32

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.
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.