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Assume you have an action:
$S[q] = \int L(q, \dot q, t)$ (i.e $q$ is a function of time). (1)
Then you do a transformation on $q(t)$ such as $\sigma(q(t), a)$ where $a$ is infinetisemal and this results in:
$S[\sigma(q(t), a)] = \int L'(\sigma(q,a), \frac{d}{dt}\sigma(q,a), t)dt$. (2)

Assume that when you do such transformation, $S[q]$ and $S[\sigma(q(t), a)]$ end up in a difference by the total time derivative of some function(i.e $\frac{d}{dt}\Lambda(q,t)$.


Now here is what the article does.

we can expand (2) in Taylor series:
$S[\sigma(q(t), a)] = \int dt(L(q, \dot q, t) + a\left[\frac{\partial L'}{\partial a}\right]_{a=0} + a^2\left[\frac{\partial L'}{\partial a}\right]_{a=0} + ...)$
$S[\sigma(q(t), a)] - S[q] = \int dt(a\left[\frac{\partial L'}{\partial a}\right]_{a=0} + a^2\left[\frac{\partial L'}{\partial a}\right]_{a=0} + ...)$
left side is our assumption that actions ended up differed by total time derivative, so:
$\int \frac{d}{dt}\Lambda(q,t) = \int dt(a\left[\frac{\partial L'}{\partial a}\right]_{a=0} + a^2\left[\frac{\partial L'}{\partial a}\right]_{a=0} + ...)$


and Now the magic happens: we neglect higher orders and say: $\int \frac{d}{dt}\Lambda(q,t) = \int dt(a\left[\frac{\partial L'}{\partial a}\right]_{a=0})$ (3)

Question:

Can you explain why this is justified? It seems to me that (3) is not a strict equality, but approximation($\approx$) - you might say that since $a$ is infinitesimal, higher orders would be super small and can be neglected. Still, I am not sure about that statement, because from (3), we derive conservation law, and if (3) is an approximation, then how can conservation law derivation be legit?

In other words, if $a = b + c + ...$, can we say $a=b$ if $c$ is infinitesimally small compared to $b$ and is this justified all the time? if yes, how can this be justified in the action example I provided where conservation law is derived from it?

I am not so keen on this exact example and we can simplify the explanation by using normal functions, I just need to understand this concept of neglection, but from a mathematical viewpoint and a physics viewpoint (as seen from mathematics as well).

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    $\begingroup$ I don't think evaluating the difference perfectly is really important in classical mechanics, the point is more of what the first order variation is, and by definition this must be function of first-order variations of the functions inside. It's like with total differential of multivariable function in mathematics, the point of it is not to express changes correctly, but to express the major part of the change, which is a linear function of the parameter differentials. $\endgroup$ Commented Dec 17, 2023 at 20:59

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I will answer in a bit more general way, I hope this still clears some of the confusion.

The crucial detail is that we are dealing with infinitesimal transformations (as the blog post says). I agree that we are making an error when approximating the transformed quantity $\sigma$ with its first order series expansion, but the error can be made as small as we like.

Think of the following simple example; I am sure you would agree that the sequence $a_n = 1/n$ in $\mathbb{R}$ approaches $0$. While for any chosen $N\in\mathbb{N}$ the term $a_N = 1/N$ is not identically zero, it can be made arbitrarily close to $0$ by choosing a large enough $N$. The same argument goes for picking $a$ in your example.

A natural question arises almost immediately; why can we say anything about a 'real' (i.e. finite) transformation, if we are dealing only with infinitesimally small transformations? Assuming everything is smooth enough, one can imagine an arbitrary smooth transformation to be constituted of (infinitely) many (infinitesimally) small transformations! If we take rotations by an angle $\theta$ in 2 dimensions for example: \begin{equation} \begin{pmatrix} \cos \frac{\theta}{N} & \sin \frac{\theta}{N} \\ -\sin \frac{\theta}{N} & \cos \frac{\theta}{N} \end{pmatrix} = I + \frac{\theta}{N}\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} =: 1 + \frac{\theta}{N}J, \end{equation} where we take $N$ large enough in order to neglect higher terms (which can't play a role in any case since we'll take the limit $N\rightarrow\infty$). We call $J$ the generator of the rotation. Now we have a description of rotating for very small angles. Then $$ \lim_{N\rightarrow\infty} (1 + \frac{\theta}{N} J)^N = e^{\theta J} = 1 + \theta J + \frac{1}{2!} \theta^2 J^2 + \dots = \cos\theta + J\sin\theta = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}, $$ where the last equalities holds for any $\theta$ since the Taylor series of the exponential map converges on the whole real axis. What we see is that for a given angle of rotation $\theta$, we can describe the full rotation (RHS) by applying sequentially applying the infinitesimally small rotation (i.e. rotation for $\theta / N$) - this is the LHS with the limit expression.

So to recap, when dealing with infinitesimals one can neglect higher terms as we can always take the parameter of expansion ($a$ in your case) to be as small as we'd like in order to make the error as small as we'd like. Then we make a claim that holds for these very small (infinitesimal) transformation. We can then show that an arbitrary transformation is composed of product of many infinitesimal transformation for which we've shown that the claim holds. Finally, the claim must therefore hold for the full transformation.

This is a bit of a hand-wavy explanation and introduction to the exponential map of a Lie group and Lie algebras which often plays a crucial part in (theoretical) physics.

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    $\begingroup$ This is the answer I was expecting. I knew it had to do with lie group and algebra but since I know nothing about the topic, it's confusing. Let me ask you the following: so we have $a$ to be infinetisemal such as it's not 0, but $a^2$ ends up ALMOST 0. it's like the following thing: $\int \frac{d}{dt}\Lambda = \int L + a\frac{\partial L}{\partial a} + $ ALMOST 0(infinitely close to 0) and because of this, ALMOST 0(infinitely close to 0) is not used to calculate difference. All correct ? $\endgroup$ Commented Dec 17, 2023 at 21:56
  • $\begingroup$ One question though: do you think on the article, we would still end up with the same derivation result if we had still said: $a\frac{\partial L}{\partial a} + \frac{a^2}{2}\frac{\partial^2 L}{\partial a^2} = \frac{d}{dt}\Lambda$ ? no chance, right ? $\endgroup$ Commented Dec 17, 2023 at 21:57
  • $\begingroup$ You're not wrong, I just think that it is more (physically) beneficial to think of this whole derivation as calculating up to first order. What I like about this way of thinking is that I always imagine that given a problem, I'm calculating the best linear approximation that describes it. This basically means what you said though, second order corrections are negligible. $\endgroup$
    – mb28025
    Commented Dec 17, 2023 at 22:04
  • $\begingroup$ As far as your second comment goes; I'll be honest that I didn't really dive too deep into the derivation itself. My suggestion is that you try it yourself and see where (if) things go wrong. All of the derivations of Noether's theorem I've seen and done were using first order results only. $\endgroup$
    – mb28025
    Commented Dec 17, 2023 at 22:06
  • $\begingroup$ Thanks so much. I have to think about this tomorrow much more. I will let you know here how it goes. You rock ! $\endgroup$ Commented Dec 17, 2023 at 22:12

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