I Taylor expansion about lagrangian in noether

AI Thread Summary
The discussion centers on the mathematical derivation of Noether's theorem, particularly how Lagrangians transform under infinitesimal transformations. It highlights that when Lagrangians differ by a total time derivative of a function, the first-order Taylor expansion suffices to establish equivalence between transformed and original Lagrangians. The key point is that the first-order term leads to a conservation law derived from the Euler-Lagrange equations. The conversation also touches on the possibility of extending these concepts to more general transformations, maintaining the symmetry condition and resulting conservation laws. Understanding these transformations is crucial for grasping the implications of Noether's theorem in physics.
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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 ?
 
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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.
 
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