- #1

Stalafin

- 21

- 0

The first:

The starting point of a symmetry operations is the following expansion:

[tex]f(r+a) = f(r) + \left(a\cdot\frac{\partial}{\partial r}\right) f(r) + \frac{1}{2} \left(a\cdot\frac{\partial}{\partial r}\right)^2 f(r) + \frac{1}{3!} \left(a\cdot\frac{\partial}{\partial r}\right)^2 f(r) + \ldots[/tex]

The second:

This one comes from Landau's book, first chapter, fourth section:

Given the equation [itex] L = L(v^{\prime 2}) [/itex] where [itex]\boldsymbol{v}^\prime = \boldsymbol{v} + \boldsymbol{\varepsilon}[/itex] and [itex]\boldsymbol{v}^2 = v^2[/itex] s.t. [itex] L(v^{\prime 2}) = L(v^2 + 2\boldsymbol{v}\boldsymbol{\varepsilon} + \boldsymbol{\varepsilon}^2)[/itex]: why does expanding in powers of [itex]\boldsymbol{\varepsilon}[/itex] and neglecting terms above first order lead to:

[tex]

L(v^{\prime 2}) = L(v^2)+\frac{\partial L}{\partial v^2} 2\boldsymbol{v}\cdot \boldsymbol{\varepsilon}

[/tex]

Some insight into why this is would be greatly appreciated. :-)