- #1
qoqosz
- 19
- 0
According to Landau textbook:
Having two inertial frames K and K' moving with velocities [tex]\vec{v}[/tex] and [tex]\vec{v'}=\vec{v} + \vec{\epsilon}[/tex] where [tex]\vec{\epsilon}[/tex] is an infinitesimal. We have [tex]L' = L(v'^2) = L (v^2 + 2 \vec{v} \cdot \vec{\epsilon} + \epsilon^2)[/tex]. Expanding this expression in powers of [tex]\vec{\epsilon}[/tex] and neglecting terms above first order, we obtain: [tex]L(v'^2) = L(v^2) + \frac{\partial L}{\partial v^2} 2 \vec{v} \cdot \vec{\epsilon}[/tex].
I simply don't understand how the second term on the right of the equation is of that form, i.e. why we differentiate with respect to v^2?
According to my calculations it should look like:
[tex]L(v'^2) \approx L(v^2) + \frac{\partial L (v^2)}{\partial \vec{\epsilon}} \vec{\epsilon} \cdot (2 \vec{v} + \vec{\epsilon} ) \approx L(v^2) + \frac{\partial L (v^2)}{\partial \vec{\epsilon}}2 \vec{\epsilon} \cdot \vec{v}[/tex]
Having two inertial frames K and K' moving with velocities [tex]\vec{v}[/tex] and [tex]\vec{v'}=\vec{v} + \vec{\epsilon}[/tex] where [tex]\vec{\epsilon}[/tex] is an infinitesimal. We have [tex]L' = L(v'^2) = L (v^2 + 2 \vec{v} \cdot \vec{\epsilon} + \epsilon^2)[/tex]. Expanding this expression in powers of [tex]\vec{\epsilon}[/tex] and neglecting terms above first order, we obtain: [tex]L(v'^2) = L(v^2) + \frac{\partial L}{\partial v^2} 2 \vec{v} \cdot \vec{\epsilon}[/tex].
I simply don't understand how the second term on the right of the equation is of that form, i.e. why we differentiate with respect to v^2?
According to my calculations it should look like:
[tex]L(v'^2) \approx L(v^2) + \frac{\partial L (v^2)}{\partial \vec{\epsilon}} \vec{\epsilon} \cdot (2 \vec{v} + \vec{\epsilon} ) \approx L(v^2) + \frac{\partial L (v^2)}{\partial \vec{\epsilon}}2 \vec{\epsilon} \cdot \vec{v}[/tex]