Undergrad Taylor expansion about lagrangian in noether

[gionole]

I was studying a derivation of noether's theorem mathematically and something struck my eyes.

Suppose you have $L(q, \dot q, t)$ and you transform it and get $L' = L(\sigma(q, a), \frac{d}{dt}\sigma(q,a), t)$. $\sigma$ is a transformation function for $q$

Let's represent $L'$ by taylor around point 0, which gives us:

$L(q, \dot q) + a\frac{\partial L}{\partial a}\Bigr|_{a=0} + a^2\frac{\partial^2 L}{\partial a^2}\Bigr|_{a=0} + ...$

Now, here is the tricky part: If lagrangians(before and after transformation) are differed by total time derivative of some function, we can say that:
$\frac{\partial L}{\partial a}\Bigr|_{a=0} = \frac{d}{dt}\Lambda$

Note that first order from taylor turned out to be enough, even though taylor only with first order is not the exact(100%) approximation of any $L$. So it turns out that we use: $a\frac{\partial L}{\partial a}\Bigr|_{a=0} = a\frac{d}{dt}\Lambda$ instead of $a\frac{\partial L}{\partial a}\Bigr|_{a=0} + a^2\frac{\partial^2 L}{\partial a^2}\Bigr|_{a=0} + ... = \frac{d}{dt}\Lambda$

I am told that we can do this because $a$ is infinetisemal transformation, but still don't get why this works. I know I need to know lie theory to understand this, but isn't there really any other way to somehow grasp it without bunch of math ?

 

[vanhees71]

You just need that $L'$ and $L$ should be equivalent Lagrangians, i.e., that there exists a function $\Omega(q,t)$ such that
$$L'[\sigma(q,a),\dot{\sigma}(q,a),t]=L(q,\dot{q},t]+\dot{\Omega}(q,t),$$
where the dots mean total derivatives (symmetry condition).

For an infinitesimal transformation, i.e., $q'=q+\epsilon \sigma(q,t)$ you can make the ansatz $\Omega(q,t)=-\epsilon \tilde{\Omega}(q,t)$ and expand the LHS of the symmetry condition to first order in $\epsilon$, which then leads to a conservation law for the solutions of the Euler-Lagrange equations.

BTW: That's only the most simple special form of the transformations considered in Noether's theorem. You can extend it to more general transformations,
$$t'=t+\epsilon \Theta(q,\dot{q},t), \quad q'=q+\epsilon Q(q,\dot{q},t).$$
Then
$$\frac{\mathrm{d}}{\mathrm{d} t'} L[q+\epsilon Q,\dot{q}+\epsilon \dot{Q},t+\epsilon \Theta(q,\dot{q},t)]=L(q,\dot{q},t) -\epsilon \dot{\tilde{\Omega}}(q,t).$$
Now again you can expand everything to 1st order in $\epsilon$ to get the symmetry condition, and then there follows again a conservation law for the solutions of the Euler-Lagrange equations.