In Landau & Liftshitz’s book p.5, they go ahead and writes down lagrangians for 2 different inertial frames. They say that Lagrangian is a function of $v^2$.
So in one frame, we got $L(v^2)$.
In another frame (the speed relative to the first frame is $\epsilon$ so its actual speed is $v+\epsilon$
So lagrangian for the second reference frame is $L(v^2 + 2v\epsilon + \epsilon^2)$.
Then, the book proceeds to mention that we need to expand this in powers of $\epsilon$ and then neglect the terms above first order, we obtain:
$L(v^2) + \frac{\partial L}{\partial v^2}2v\epsilon$ (This is what I don't get).
We know taylor is given by: $p(x) = f(a) + f'(a)(x-a) + ....$
I tried doing this in respect to $\epsilon$. We know $\epsilon$ is super small, so our taylor is $f(0) + f'(a)(x-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}(2v+2\epsilon)|(\epsilon=0)(\epsilon-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}2v\epsilon$ (note that I don't have $\frac{\partial L}{\partial v^2}$)
If I try doing taylor with respect to $v^2$, then I don't get $L(v^2)$ as the first part because $f(a) != L(v^2)$
Where am I making a mistake?
Update
A couple of things that confuse me.
- $\frac{d}{d\epsilon}L(v^2 + 2v\epsilon + \epsilon^2)|_{\epsilon = 0} = \frac{d}{d\epsilon}L(x)|_{x = v^{2}}$ If we imagined $x = v^2 + 2v\epsilon + \epsilon^2$, why do we say $|x=v^2$ only ?
- $[\frac{d}{dx} L(x)]|_{x = v^2} = \frac{d}{dv^2} L(v^2)$ (here the left pat says differentiate L with x and then put $v^2$ instead of $x$. how is this the same as the right part ?
