# Lagrangian for a free particle

1. Feb 2, 2014

### Bryson

Hello, this is probably one of those shoot yourself in the foot type questions.

I am going through Landau & Lifshits CM for fun. On page 7 I do not understand this step:

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

where $v' = v + \epsilon$. He then expands the expression in powers of $\epsilon$ (neglecting higher order terms) to get:

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

How did he arrive here? What expansion did he use? Taylor expansion?

Thanks for any help or comments!

2. Feb 5, 2014

### Avodyne

Yes, Taylor expansion. Higher-order terms are neglected because they want to determine the infinitesimal change in L that follows from an infinitesimal change in v. (I'm guessing here because I don't have the book, but there's no other reason to do this calculation. I presume that the final result will be the Euler-Lagrange equations.)