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Hello I have a quick question on what I have been reading in Landau & Lifshitz's book on classical mechanics. I am in the very beginning of the book and I am having trouble with his derivation on the Lagrangian of a free particle. At the very top of page 7, he equates two Lagrangian by

$$\mathrm{L}'=\mathrm{L}(v'^2)=\mathrm{L}(v^2+2v \cdot \epsilon+ \epsilon^2).$$

Then he goes to to say that you can expand this in powers of $ \epsilon$ and neglect all terms above the first order to get

$$\mathrm{L}(v'^2)=\mathrm{L}(v^2)+\frac{\partial \mathrm{L}}{\partial v^2}2v \cdot \epsilon.$$

How do you expand in terms of $\epsilon$ to get this result? This Lagrangian seems to me to be a function of three variables and I am looking at the multivariable Taylor expansion and I am not seeing how Landau could get his result.

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Landau defines $v'\equiv v+\epsilon$. Then

$$ L(v'^2)= L[ (v+\epsilon)^2] = L(v^2+2v\cdot\epsilon+\epsilon^2)\sim L(v^2+2v\cdot \epsilon) $$

In the last step we have omited the second order terms in $\epsilon$.

So we have $L(v^2+2v\cdot\epsilon)$ which is a function like $f(a+x)$. We expand it as

$$ f(a+x)\sim f(a)+f'(a)x $$

at first order. In our case, we have $a=v^2$ and $x=2v\cdot\epsilon$, so

$$ L(v'^2)\sim L(v^2 + 2v\cdot\epsilon) \sim L(v^2) + \frac{\partial L}{\partial v^2} 2v\cdot\epsilon$$

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  • $\begingroup$ But since you defined x=2v*e, shouldn't the derivative of f be taken with respect to that? You seemed to have taken it with respect to v^2 instead, which was your a. (I'm talking about what happened in the very last line). $\endgroup$ – Physics Llama Dec 10 '15 at 12:45
  • $\begingroup$ In fact, $L$ is a function of $v'^2$, and you are deriving wrt this variable, but then you are evaluating $v'^2=v^2$, as in every Taylor expansion. $\endgroup$ – dpravos Dec 15 '15 at 9:30
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Consider $L(v'^2)$ as a function of the variable $\epsilon$ it is

\begin{equation} L(v'^2)=F(\epsilon) \end{equation}

then use the conventional Taylor expansion around $\epsilon=0$

\begin{equation} F(\epsilon)=F(0)+\frac{dF(0)}{d\epsilon}\cdot\epsilon+\mathcal{O}(\epsilon^2) \end{equation}

then using the fact that $dF/d\epsilon=(\partial F/\partial v'^2)(dv'^2/d\epsilon)=(\partial F/\partial v'^2)2v'$ and that $v'$ reduces to $v$ when you evaluate at $\epsilon=0$, you get

\begin{equation} L(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}2v\cdot\epsilon+\mathcal{O}(\epsilon^2) \end{equation}

which is the desired relation.

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