Noether's Theorem Explained (Part 3/6) - Angular Momentum and Rotations 

That's a good question. When we throw out O(ε^2) terms in intermediate steps, we are implicitly assuming that they aren't somehow "square rooted" later on, which would produce an O(ε) term in the final answer. You do always have to be careful when throwing out higher order terms for this reason. Thankfully, we can check that throwing out the O(ε^2) is justified here. The term in question is V( sqrt(x^2 + y^2) ). Under an infinitesimal rotation by the angle ε, x^2 + y^2 does not change to the first order in ε. However, it does change to the second order in epsilon. At 16:14 in the video, you can see that the second order term is ε^2(x^2 + y^2). Now, what is V(sqrt(x^2 + y^2 + ε^2(x^2 + y^2) ) ) to the first non-vanishing order in ε? Well, note that we can rewrite this as V( sqrt(x^2 + y^2) sqrt(1 + ε^2) ). Next, note that sqrt(1 + ε^2) ≈ 1 + (1/2)ε^2 (because the first derivative of sqrt(1 + z) at z = 0 is 1/2). Therefore, we can see that V(sqrt(x^2 + y^2 + ε^2(x^2 + y^2) ) ) ≈ V(sqrt(x^2 + y^2) ) + (1/2)ε^2 V(sqrt(x^2 + y^2) ). So, in the end, yes, we do see that the O(ε^2) term inside the square root doesn't actually "get square rooted" to become an O(ε) term, as feared, but will result in an O(ε^2) once all is said and done. This is because the ε^2 term is actually being expanded around 1, not 0! There's a big difference between sqrt(1 + ε^2) and sqrt(ε^2). Here is another way you should know that V shouldn't change with an O(ε) term. Note that we could have just as easily written the potential energy as V(x^2 + y^2), instead of V(sqrt(x^2 + y^2)), because V is an arbitrary function. There would have been no physical difference between these two choices, only a notational difference. It would have been weird if simply writing V(sqrt(...)) instead of V(...) mattered when V is an arbitrary function anyway.

​@@noahexplainsphysics Thank you for the in-depth answer. I wouldn't have found it myself, but I totally buy it.

@@noahexplainsphysics Okay, but we do end up with a constraint on the form of V. e.g. if V = sqrt then those terms can't be discarded. Is that a correct interpretation? If so, nteresting that the form of our potential has an impact on conservation. (Love the vids, BTW. Moar plz.)

@@stephenmcateer That will only cause an issue at 0, because sqrt(x + ε) = sqrt(x) + ε/2sqrt(x) + ... . Notice that the O(ε) term is perfectly fine except in the case that x = 0. The derivative of sqrt(x) is finite, and thus is able to be taylor expanded, at any point except x=0 where its derivative blows up.

One such example is the hidden SO(4) symmetry in the Kepler problem/Hydrogen atom. A second example is rotating position and velocity into each other in the harmonic oscillator. (For an N dimensional harmonic oscillator, this is U(N) symmetry.) A third example is the SL(2,R) symmetry of the 'conformal' particle, which has V(x) = a/x^2.

Because you consider de motion to start at t1 and finish at t2, so epsilon, which is a tiny variation in motion does not make sense to be other than zero at the beginning and end of the motion, because you would start/end up in some other place ! I guess you already figured it out since your comment was 2 months ago, but I suggest learning about the principle of least action to understand variation in the motion !