The trajectory is not straight because Feynman is imagining an arbitrary system, it can be a many-body system, even a field, and the trajectory can be complicated. The symmetry relates the bulk trajectory to the bulk shifted trajectory, and the stationary property means that the total action is unchanged by the infinitesimal deformation, so you can conclude that the top little bit is equal to the bottom little bit of action.
Formal equivalent
To make Feynman's argument formally clear, let's consider the action for two particles interacting by a translationally invariant force-law:
S=∫m12˙x12+m22˙x22−V(x1−x2)dt
The minimizing trajectory is the pair x1(t),x2(t). Now consider making an infinitesimal variation of the path of Feynman's form, taking the trajectory-pair x1,x2 to x1(t)+ϵθ(t),x2(t)+ϵθ(t), where ϵ is infinitesimal, and θ(t)=1 for t0<t<t1 and θ(t)=0 elsewhere. This is translating the system between times t0 and t1, and it produces two "horizontal lines" at t1 and t2 (in the higher dimensional configuration space of the two particles).
As Feynman says, the action of this perturbed path is equal to the original action, to lowest order, because the true path is an extremum, and this is an infinitesimal variation. But aside from the two infinite-velocity "kicks" at time t0 and t1, the action is also exactly equal to the old action, by translation invariance. So you can conclude that the action contribution of the two velocity-kicks have to cancel out, i.e., the action of the horizontal segments at t0 and t1 are equal, and something is conserved.
To find what that something is, the formal action of these horizontal segments is found formally by writing the velocity perturbation δ˙x--- the change in velocity is formally ϵδ(t−t0)−ϵδ(t−t1), i.e. the velocity of each particle gets two infinitesimal delta function kicks.
The formal variational change in L in response to this infinitesimal variation of the velocity is ∂L∂˙x1δ˙x1+∂L∂˙x2δ˙x2, which is ϵ(P1(t)+P2(t))(δ(t−t0)−δ(t−t1)), i.e., the total momentum times the variation in velocity. But since the variation in velocity is just the two delta functions, with opposite sign, the total variation, which is zero, ends up saying that the total momentum is equal at the beginning and at the end, that is, that the total momentum is conserved. This is Feynman's argument made formal, and you can see that it is formally correct and equivalent to other approaches.
The variation in S is the integral of the variation of L: ∫(P1(t)+P2(t))(δ(t−t0)−δ(t−t1))dt or PT(t1)−PT(t0) where PT is the total momentum. Setting this to zero means PT is conserved.
The intuition problem with this formal argument is simply the use of variational derivatives and delta-function variations of the velocity. Variational derivatives are counterintuitive, because, given a functional F(x(t)) they are formally "defined" in physics as follows:
δFδx(t)(t0)=F(x(t)+ϵδ(t−t0))−F(x(t))ϵ
This definition is patent nonsense (although useful heuristically), because no matter how small ϵ is imagined to be, the δ function is infinite at t0, and so the variation cannot possibly ever be truly infinitesimal.
To define the variational derivative mathematically properly requires the following--- for any ϵ(t), the infinitesimal variation in F(x(t)) is linear to first order, so it must be a linear (possibly distributional) kernel integrated against ϵ:
F(x(t)+ϵ(t))=F(x(t))+∫K(t′)ϵ(t′)dt′+o(ϵ)
where the functional is differentiable when there is a K making the above work, and by definition, the kernel K is the variational derivative:
δFδx(t)(ti)=K(ti)
Then you can see why the formal definition works, because when you formally plug in ϵ(t)=δ(t−t0), you do the integral for the variation, and you pick out K(t0). This is why the formal definition, despite being nonsense, is useful for pictures, because it is sort of what you are doing. But in reality, if you want things to be infinitesimal, you have to imagine the delta function is regulated to a skinny bump, and ϵ is small enough to make the bump infinitesimal.
In this case, the confusions about "infinite velocity" from the horizontal kicks is entirely due to the issue of taking functional derivatives with delta functions, instead of doing a smooth variation, and substituting delta functions only at the end. But this is a confusion which disappears upon familiarity with functional derivatives, and glossing over this annoying subtlety is actually a nice feature of Feynman's argument, because you need to fill in this subtlety yourself, and this is extremely good training for students who are trying to gain intuitive familiarity with functional derivatives.
The upshot of all this is that Feynman's argument is completely 100% correct, but to see this, it relies on you being comfortable with the notion of functional derivative, and with the idea of an "infinitesimal" delta function variation of the velocity. You can generalize the argument to any symmetry, and you recover the correct formula for the Noether current, and I personally find it easiest to derive the Noether current in this way.
The argument appears in the non-popular physics literature after 1964 in various places in disguise. One famous way people prove Noether's theorem is by allowing the symmetry parameter to vary with time (or with time and position in field theory), and then integrating by parts, noting that total variation is zero, and the variation is proportional to a derivative. While this looks formally different from Feynman's 1964 proof, it is completely identical when push comes to shove, it is just folding in the functional derivative issues and making them disappear, so that it is conceptually doesn't require understanding δ function variations. This is the textbook proof most often presented to students today, I read it in one of Hawking's articles. But it's really Feynman's 1964 argument said in slightly more and less accessible language.
The beauty of Feynman's argument is that it is simultanously more and less accessible. It is more accessible to the lay-public in that the intuition is immediate and apparent when you have no mathematical training. It is less accessible because the variational derivative issue means that it looks like nonsense when you have a little bit of mathematical training. The beauty of it is that once you sort out the issues that come up on second glance, it becomes correct again, and you have learned two things for the price of one very simple diagram and very simple argument.
This argument by Feynman is, by far, my favorite popular science, because it is the only case in history I know of where a result was presented at the popular level only to become standard at the professional level later. Needless to say, the flow of ideas usually goes only in the opposite direction. In addition, this argument is very timeless and beautiful, and it's superficial apparent wrongness at an intermediate level is part of what makes it so beautiful.
Original Answer
(I originally didn't write all the above, as I decided just to replace the delta functions in Feynman's argument by smoothed out delta function kicks, and then I know the argument goes through. But this turned out to be confusing, so I expanded the answer)
But you can understand Feynman's argument more cheaply by just regulating the delta-functions to quick infinitesimally small impulses of velocity. If you do this, then using horizontal lines is a little misleading, because when you make the argument precise in this way, the lines are only infinitesimally different in velocity, so they are nearly parallel, they just slide out a little bit, then back at the end. But this is hard to draw.
To regulate the argument, replace the horizontal lines with short lines, of height ϵ and horizontal width δ. The velocity along the horizontal segment is δ/ϵ, and the extra action along the little segment at the beginning is ∂S∂˙xδϵ, which is pv. You are free to adjust ϵ and δ independently, so you can make v small and the segment short at the same time. In this limit, the lines are nearly vertical, but this doesn't look good in a picture.
The result is that the canonical momentum corresponding to the coordinate being translated is conserved. In a many particle system, the sum of the momenta over all the particles is the conserved quantity, since the action is added up over the particles being translated.
For any motion of the coordinate, the little-path action is pδx in the limit of small δx. So for time translation, you get the little-action p˙x−L, for the endpoints, by moving the path a little bit up in time. Moving the points up, you move them over by their velocity, so p˙x, while you are removing a little bit of action because the range of integration is shifted, so there is a second contribution which is Lδt. This gives the Hamiltonian, and you get the law of conservation of energy.
Feynman's argument is most often presented by making the infinitesimal translation depend on time, and integrating by parts. This is completely equivalent, because you can imagine the little kicks being distributed along the time-axis uniformly. In the important special case of an action quadratic in the velocity, the actual change in action does not care about the slope of the line, and you can make the lines horizontal by taking the limit of δ→0, ϵ→0, δϵ→∞, and the change in action along the (now horizontal) paths is the same as if the translation is infinitesimal.
This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Ron Maimon