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I'm hung up on page 7 of Landau & Lifshitz Course on Mechanics. They claim,

$$L(v'^2) = L(v^2)+\frac{\partial L}{\partial v^2}2\textbf v\cdot \epsilon \tag{p.7}$$ The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $\textbf v$.

I can understand the Taylor expansion to get to this equation, but I do not understand why this must be a linear function of $\textbf v$.

If I understand correctly, we need to find a function$f(q, t)$ such that:

$$\frac{d}{dt}(f(q,t)) = \frac{\partial L}{\partial v^2}2\textbf v\cdot \epsilon$$

If $f(q,t) = q^2$, would the total time derivative not include a nonlinear function of $\textbf v$? Is there some restriction imposed upon $f$ such that it must also be a linear function of $q$?

Qmechanic
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Kyle Tennison
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1 Answers1

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The Taylor expansion is:

$f(x+\delta)=f(x)+f'(x)\delta+...$

Here $x=v^2$, and you know that

$v^2+\delta=(\vec{v}+\vec{\epsilon})^2=v^2+2\vec{v}.\vec{\epsilon}+\epsilon^2$

thus,

$\delta=2\vec{v}.\vec{\epsilon}+\epsilon^2$

discarding terms of order $\epsilon^2$, then you get the expression in the book.

Pato Galmarini
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