Noether's Theorem Explained (Part 2/6) - Momentum and Spatial Translations 

The invariance does matter but it may not be apparent how on a first pass. In the fourth video in this series I explain why this procedure works in general (and why the invariance matters) in more detail. In fact, if the Lagrangian weren't invariant under spatial translation then the total momentum would not be conserved but would in fact change in time. (There would be a net force.) I demonstrate this for angular momentum and rotations in the next video.

The key idea is that E(t)≠0 for any t. It is true that E(t1) = E(t2) = 0, but if t≠t1 and t≠t2, E(t) does not have to be zero. In fact, the point is that we can choose E(t) to be any function we want as long as E(t1) = E(t2) = 0. If we have that the integral of E(t)*f(t) is 0 no matter what E(t) is, then we must have that f(t) = 0 for all t. The reason for this is that, if there were some t0 for which f(t0)≠0, then we could choose E(t) to be a little "bump" at t0, so that the integral of E(t)*f(t) would not be zero. This is the most important trick of Lagrangian mechanics. For more information, watch my video on the Principle of Least Action, where I use that trick to derive Newton's law F = ma.

The action S is a function of a path. So you plug in a path, say x(t), and you get the action of the path S[x(t)]. In the beginning of the video I show that the action doesn't change under a translation, so S[x(t)] = S[x(t) + a], where here "a" is a constant vector added to the path. Note that this is true for any arbitrary path x(t). Even if x(t) is some random squiggly path that doesn't minimize the action, x(t) and x(t)+a have the same action. Now, xbar(t) is the "true path," the one that minimizes the action. When we make a tiny variation and expand out the equation δS = 0, we are expanding around the true path xbar(t), not any old arbitrary path x(t). So to reiterate: the action S has a translation symmetry after translating any path x(t) by a constant vector, but the equation δS = 0 only holds for variations around the true path xbar(t).

@@noahexplainsphysics ah. That makes more sense. Thank you!

I am not entirely sure nor qualified to comment but it may be because a translation of the system by a function ε would affect every particle identically?

Well in that general case while the kinectic energy will still be preserved, the potential energy will change, remember the potential energy given by V depends upon the distance of the particles so if you pick epsilon_1 different from epsilon_2 the potential energy will change in some time intervals, and while in some lucky cases the time integral of V can still be conserved, this is not the general case. Basically making them equal proves there is a continuos 1 parameter group acting as a symmetry, while if your case were also true you would have a continuos 2 parameter group acting as symmetry which could be thought of as two continuos 1 parameter groups acting, then by Noether's theorem we woild have two preserved quantities,that is, another one besides the total momentum. This part I think is not possible because then you have three quantities preserved (including the lagrangian), but this is only possible if the third quantity is linear combination of the first two

"The term "stationary" means that any variation of \bar{x}(t) doesn't change S to the first order in epsilon, i.e. 𝛿S[ \bar{x}(t)] = 0. In other words, if \bar{x}(t) satisfies the equation of motion, then it makes S "stationary" with respect to small variations.

@@noahexplainsphysics thanks for the clarification! Your efford is really appreciated! If you have a link for donations or similar, you could drop it in the video description

Fix a starting point, an ending point, a starting time t_1 and an ending time t_2. The Principle of Least Action states that the path which a particle takes from the start point at t_1 to the end point at t_2 is the one which minimizes the action S. Therefore, in order to use the Principle of Least Action to get the equations of motion, we must consider all little tiny variations around the path which, crucially, have the same start and end points! (This is because the action is only minimized for the set of all paths which start and end at the same points.) That is, we consider all paths x(t) + ε(t) where ε(t_1) = ε(t_2) = 0. So it is not an "assumption" as much as a requirement necessary to use the Principle of Least Action. If you want more details, I discuss this more in my video on the Principle of Least Action.

@@noahexplainsphysics Your answer is great. It confirms what I thought as well. You are the best and thanks

It seems sufficient to assume only epsilon(t_1) = epsilon(t_2) without assuming that they’re zero. Isn’t it correct to only assume this: Given that we’re exploiting translational symmetry, we also want to translate the start and end points of our path.

Usually when dealing with Lagrangian mechanics, we only think about infinitesimal variations. This is because the action of the true path doesn't change to the first order in a tiny variation, which is how you can prove statements in the subject. There's not really a way to use finite (not infinitesimal) symmetries in Lagrangian mechanics to prove stuff. As for why the translation is time dependent when translations aren't usually time dependent, (and I admit the logic is a bit subtle) it's because this is how the trick works. The Lagrangian is invariant under a constant (not time dependent) translation. However, if you instead consider a time dependent tiny translation, and consider how the action S changes to the first order, you can use that 𝛿S = 0 is true for the actual physical path and show this implies the conservation of momentum. So the time dependent tiny translations should be thought of as a tool to derive the conservation law. I explain this logic more explicitly in my fourth video in this series, which might clear up some of your confusions.

@@noahexplainsphysics Thank you so much !

That is a good question which I do not address in this video but I address in parts 3 and 4 of this series. It turns out that if you write a Lagrangian which does not have a space translation symmetry, namely if the change in the Lagrangian under a tiny translation parameterized by ɛ is 𝛿L = ɛA where A ≠ 0, then the linear momentum P isn't conserved in time, and in fact dP/dt = A. This means that A is a net force. So yes, when we have a symmetry we get that our quantity is conserved, and when we don't have a symmetry we get that our quantity isn't conserved.

@@noahexplainsphysics Ah -cool. I will force ahead then. Guess I was getting greedy. :-)