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 $v$

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 $v$

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 $v$

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 $v$

1. On page 4 of Landau & Lifshitz's Mechanics he says

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

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

How do we get this?

2. Also, what is the meaning of 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 $\bf v$

3. 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 $v$

I don't feel so dumb anymore knowing that so many people have this question! But what about the second part of my question?

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 $v$