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In Landau-Lifschitz, following expansion is given, We have, $$L(v'^2)~=~L(v^2+2\textbf{v}\cdot\epsilon+\epsilon ^2)$$ expanding this in powers of $\epsilon$ and neglecting powers of higher order, $$L(v'^2)~\approx~L(v^2)+\frac{\partial L}{\partial (v^2)}2\textbf{v}\cdot\epsilon$$

$L$ is Lagrangian. and $\textbf{v}'=\textbf{v}+\epsilon$. I am unable to follow what expansion this exactly is. It does not look like Taylor expansion.

Qmechanic
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user59756
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1 Answers1

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But it's a Taylor expansion:

$$ L(x_0+x)=L(x_0)+\frac{\partial L}{\partial x}(x_0)\cdot x $$

Now set $x_0=v'^2$, $x=2 \vec v \cdot \vec \epsilon + \epsilon^2$. Then neglecting powers of second order in epsilon leads to the desired result.

pressure
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