# What's the interpretation of Feynman's picture proof of Noether's Theorem?

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On pp 103 - 105 of The Character of Physical Law, Feynman draws this diagram to demonstrate that invariance under spatial translation leads to conservation of momentum:

To paraphrase Feynman's argument (if I understand it correctly), a particle's trajectory is the path AB. Space is horizontal; time vertical.

Because of spatial translation symmetry, the path CD has the same action as AB. Because AB has stationary action, ACDB has the same action as well.

That means the action of AC and BD are the same. (Note that they are traversed in opposite directions on the path ACDB.) This is a conserved quantity, and it turns out to be the momentum.

My question is about the meaning of the action of AC and BD. These paths aren't physical trajectories; they represent infinite velocity. I tried thinking of the trajectory as if the velocity, as a function of time, has two delta functions in it. However, because the Lagrangian depends on $v^2$, I think this leads to infinite action for the horizontal segments.

Mathematically, I see that the symmetry here implies $\frac{\partial L}{\partial x} = 0$. Least action implies $\frac{\partial L}{\partial x} = \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{x}}$. Combining these shows that the momentum $\frac{\partial L}{\partial \dot{x}}$ is constant, but I don't quite understand the connection to the picture.

Also, Feynman doesn't describe how he knows the conserved quantity is the momentum. Is there a way to get this from the picture? Finally, if momentum is being conserved, why isn't the trajectory a straight line?

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Mark Eichenlaub
asked Jan 22, 2012
For those of you who don't have the book, let me mention that Feynman makes a similar argument approximately 50 minutes into this Youtube video. Noether's theorem is covered in 45:25-51:27.

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Qmechanic
@Qmechanic Thanks. Google books didn't have an electronic copy of this book to link to.

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Mark Eichenlaub
awesome question!

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Physiks lover

## 3 Answers

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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:

$$S = \int {m_1\over 2} {\dot x_1}^2 + {m_2\over 2} {\dot x_2}^2 - V(x_1 - x_2) dt$$

The minimizing trajectory is the pair $x_1(t),x_2(t)$. Now consider making an infinitesimal variation of the path of Feynman's form, taking the trajectory-pair $x_1,x_2$ to $x_1(t) + \epsilon \theta(t) , x_2(t) + \epsilon\theta(t)$, where $\epsilon$ is infinitesimal, and $\theta(t)=1$ for $t_0<t<t_1$ and $\theta(t)=0$ elsewhere. This is translating the system between times $t_0$ and $t_1$, and it produces two "horizontal lines" at $t_1$ and $t_2$ (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 $t_0$ and $t_1$, 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 $t_0$ and $t_1$ 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 $\delta \dot{x}$--- the change in velocity is formally $\epsilon \delta(t-t_0) - \epsilon\delta(t-t_1)$, 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 ${\partial L \over \partial \dot{x_1}} \delta \dot{x_1} + {\partial L\over \partial \dot{x_2}} \delta\dot{x_2}$, which is $\epsilon (P_1(t) + P_2(t))(\delta(t-t_0) - \delta(t-t_1))$, 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: $\int (P_1(t) + P_2(t))(\delta(t-t_0) - \delta(t-t_1)) dt$ or $P_T(t_1) - P_T(t_0)$ where $P_T$ is the total momentum. Setting this to zero means $P_T$ 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:

$${\delta F \over \delta x(t)}(t_0) = {F(x(t) + \epsilon\delta(t-t_0)) - F(x(t)) \over \epsilon}$$

This definition is patent nonsense (although useful heuristically), because no matter how small $\epsilon$ is imagined to be, the $\delta$ function is infinite at $t_0$, and so the variation cannot possibly ever be truly infinitesimal.

To define the variational derivative mathematically properly requires the following--- for any $\epsilon(t)$, the infinitesimal variation in F(x(t)) is linear to first order, so it must be a linear (possibly distributional) kernel integrated against $\epsilon$:

$$F(x(t) + \epsilon(t)) = F(x(t)) + \int K(t') \epsilon(t') dt' + o(\epsilon)$$

where the functional is differentiable when there is a K making the above work, and by definition, the kernel K is the variational derivative:

$${\delta F \over \delta x(t)}(t_i) = K(t_i)$$

Then you can see why the formal definition works, because when you formally plug in $\epsilon(t) = \delta(t-t_0)$, you do the integral for the variation, and you pick out $K(t_0)$. 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 $\epsilon$ 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 $\delta$ 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 $\epsilon$ and horizontal width $\delta$. The velocity along the horizontal segment is $\delta/\epsilon$, and the extra action along the little segment at the beginning is ${\partial S\over \partial \dot{x}}{\delta\over \epsilon}$, which is $pv$. You are free to adjust $\epsilon$ and $\delta$ 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 $p\delta x$ in the limit of small $\delta x$. So for time translation, you get the little-action $p\dot{x} - L$, 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 $p \dot{x}$, while you are removing a little bit of action because the range of integration is shifted, so there is a second contribution which is $L\delta t$. 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 $\delta\rightarrow 0$, $\epsilon\rightarrow 0$, ${\delta\over \epsilon}\rightarrow\infty$, 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
answered Jan 23, 2012 by (7,720 points)
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1) Off-shell vs. on-shell action. What may cause some confusion is that Noether's theorem in its original formulation only refers to the off-shell action functional

$$\tag{1} I[q;t_i,t_f]~:=~ \int_{t_i}^{t_f}\! {\rm d}t \ L(q(t),\dot{q}(t),t),$$

while Feynman's proof [1]$^1$ mostly is referring to the Dirichlet on-shell action function

$$\tag{2} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl};t_i,t_f],$$

where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the extremal/classical path, which satisfies the equation of motion (e.o.m.)

$$\tag{3}\frac{\delta I}{\delta q} ~:=~\frac{\partial L}{\partial q} - \frac{ d}{dt} \frac{\partial L}{\partial \dot{q}}~\approx~ 0,$$

with the Dirichlet boundary conditions

$$\tag{4} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f.$$

See also e.g. this Phys.SE answer. [Here the $\approx$ symbol means equality modulo e.o.m. The words on-shell and off-shell refer to whether e.o.m. are satisfied or not.]

2) Noether's theorem. Let us recall the setting of Noether's theorem. The off-shell action is assumed to be invariant

$$\tag{5} I[q^{\prime};t^{\prime}_i,t^{\prime}_f]~=~ I[q;t_i,t_f]$$

under an infinitesimal global variation

$$\tag{6} t^{\prime}-t~=~\delta t~=~\varepsilon X(t) \qquad \text{and}\qquad q^{\prime}(t^{\prime})- q(t)~=~ \delta q(t) ~=~ \varepsilon Y(t).$$

Here $X$ is a horizontal$^2$ generator, $Y$ is a generator, and $\varepsilon$ is an infinitesimal parameter that is independent of $t$.

Noether's theorem. The off-shell symmetry (5) implies that the Noether charge $$\tag{7} Q~:=~p Y - h X$$ is conserved in time $$\tag{8} \frac{dQ}{dt}~\approx~0$$ on-shell.

Here

$$\tag{9} p~:=~\frac{\partial L}{\partial \dot{q}} \qquad \text{and}\qquad h~:=~p\dot{q}-L$$

are by definition the momentum and the energy function, respectively.

3) Assumptions. Let us assume$^3$:

1. that the Lagrangian $L(q,v,t)$ is a smooth function of its arguments $q$, $v$, and $t$.

2. that there exists a unique classical path $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ for each set $(q_f,t_f;q_i,t_i)$ of boundary values.

3. that the classical path $q_{\rm cl}$ depends smoothly on the boundary values $(q_f,t_f;q_i,t_i)$.

4) Differential ${\rm d}S$.

Lemma. The Dirichlet on-shell action function $S(q_f,t_f;q_i,t_i)$ is a smooth function of its arguments $(q_f,t_f;q_i,t_i)$. The differential is $$\tag{10} {\rm d}S(q_f,t_f;q_i,t_i) ~=~ (p_f {\rm d}q_f - h_f {\rm d}t_f) -(p_i {\rm d}q_i - h_i {\rm d}t_i),$$
or equivalently, $$\tag{11} \frac{\partial S}{\partial q_f}~=~p_f , \qquad \frac{\partial S}{\partial q_i}~=~-p_i,$$ and $$\tag{12} \frac{\partial S}{\partial t_f}~=~-h_f, \qquad \frac{\partial S}{\partial t_i}~=~h_i.$$

Proof of eq. (11):

      ^ q
|       ____________________________
|      |            q*_cl           |
|      |                            |
|      |____________________________|
|                   q_cl
|
|
|------|----------------------------|-----> t
t_i                          t_f


Fig. 1. Two neighbouring classical paths $q_{\rm cl}$ and $q^{*}_{\rm cl}$.

Consider a vertical infinitesimal variation $\delta q$ between two neighbouring classical paths $q_{\rm cl}$ and $q^{*}_{\rm cl}=q_{\rm cl}+\delta q$, cf. Fig.1. The change in the Lagrangian is

$$\tag{13} \delta L ~=~ \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} ~\stackrel{(3)+(9)}{=}~ \frac{\delta I}{\delta q} \delta q + \frac{d}{dt}(p~\delta q) ~\stackrel{(3)}{\approx}~\frac{d}{dt}(p~\delta q),$$

so that

$$\tag{14} \delta S ~\stackrel{(2)}{\approx}~\delta I ~\stackrel{(1)}{=}~ \int_{t_i}^{t_f}\! {\rm d}t ~\delta L ~\stackrel{(13)}{\approx}~[p~\delta q]_{t_i}^{t_f} ~=~p_f~\delta q_f- p_i~\delta q_i.$$

This proves eq. (11).

Proof of eq. (12):

      ^ q
|
q*_f|-------------------/
|                  /|
|                 / |
|                /  |
q_f|---------------/   |
|              /|   |
|             / |   |
|        q_cl/  |   |
|           /   |   |
q_i|----------/    |   |
|         /|    |   |
|        / |    |   |
|       /  |    |   |
q*_i|------/   |    |   |
|      |   |    |   |
|------|---|----|---|-----> t
t*_i t_i  t_f t*_f


Fig. 2. The classical path $q_{\rm cl}$.

Next consider the classical path $q_{\rm cl}$ between $(t_i,q_i)$ and $(t_f,q_f)$, cf. Fig. 2. Imagine that we infinitesimally extend both ends of the time interval $[t_i,t_f]$ to $[t^{*}_i,t^{*}_f]$, where

$$\tag{15}\delta t_i~:=~t^{*}_i - t_i \qquad\text{and}\qquad \delta t_f~:=~t^{*}_f - t_f$$

both are infinitesimally small. This induces a change of the boundary positions (4) of the fixed classical path $q_{\rm cl}$ as follows

$$\tag{16} \delta q_i~:=~ q^{*}_i - q_i~=~\dot{q}_i ~\delta t_i \qquad \text{and}\qquad \delta q_f~:=~ q^{*}_f - q_f~=~\dot{q}_f ~\delta t_f,$$

which are dictated by the end point velocities. We would now like to calculate the variation

$$S(q^{*}_f,t^{*}_f;q^{*}_i,t^{*}_i) - S(q_f,t_f;q_i,t_i) ~=~\delta S ~\stackrel{(11)}{=}~p_f \delta q_f +\frac{\partial S}{\partial t_f} \delta t_f -p_i \delta q_i + \frac{\partial S}{\partial t_i}\delta t_i$$ $$\tag{17} ~\stackrel{(16)}{=}~(p_f \dot{q}_f +\frac{\partial S}{\partial t_f}) \delta t_f -(p_i \dot{q}_i - \frac{\partial S}{\partial t_i})\delta t_i.$$

Since the new classical path is just an infinitesimal extension of the same old classical path, we may also estimate the variation as

$$\tag{18} \delta S~=~S(q^{*}_f,t^{*}_f;q_f,t_f)+S(q_i,t_i;q^{*}_i,t^{*}_i) ~=~ L_f \delta t_f - L_i \delta t_i.$$

Comparing eqs. (9), (17) and (18) yields eq. (12).

Corollary. The Dirichlet on-shell action along an infinitesimal path segment generated by the infinitesimal symmetry transformation (6) is proportional to the Noether charge $$\tag{19} S(q_i+\delta q,t_i+\delta t;q_i,t_i)~=~\varepsilon Q_i.$$

Proof of the Corollary:

$$\tag{20} S(q_i+\delta q,t_i+\delta t;q_i,t_i) ~\stackrel{(10)}{=}~p_i\delta q -h_i \delta t ~\stackrel{(6)}{=}~\varepsilon(p_i Y -h_i X) ~\stackrel{(7)}{=}~\varepsilon Q_i.$$

5) Feynman's four-point argument. We are finally ready to discuss Feynman's four-point argument.

 ^ q
|
|           A'                          B'
|            ___________________________
|           |     virtual/off-shell     |
|           |                           |
|           |                           |
|           |___________________________|
|           A    classical/on-shell     B
|
|
|------------------------------------------------> t


Fig. 3. Feynman's four points. (Note that the two horizontal and the two vertical straight ASCII lines are in general an oversimplification of the actual paths.)

We start with the on-shell action

$$\tag{21} S(A\to B)~=~I(A\to B)$$

for some classical path $q_{\rm cl}$ between two spacetime events $A$ and $B$. We then apply the infinitesimal transformation (6) to produce a virtual path $q^{\prime}$ between two infinitesimally shifted spacetime events $A^{\prime}$ and $B^{\prime}$. In turn, the virtual path $q^{\prime}$ has an off-shell action

$$\tag{22} I(A^{\prime}\to B^{\prime})~=~I(A\to B)$$

equal to the original action due to the off-shell symmetry (5).

Next we would like to consider the shifted path $A\to A^{\prime}\to B^{\prime}\to B$. Unfortunately, the two infinitesimal pieces $A\to A^{\prime}$ and $B^{\prime}\to B$ (which we will choose to be classical paths) may correspond to constant time. [The time-integration in the definition (1) of the off-shell action $I(A\to A^{\prime}\to B^{\prime}\to B)$ would not make sense in case of constant time.] In such cases we replace Feynman's four points with six points, i.e. we extend infinitesimally the original classical path $A\to B$ to a classical path $A^{*}\to B^{*}$, in such a way that the two new infinitesimal paths $A^{*}\to A^{\prime}$ and $B^{\prime}\to B^{*}$ (which we also will choose to be classical paths) do both not correspond to constant time.

 ^ q
|
|           A'                          B'
|            ____________________________
|          /|     virtual/off-shell     |\
|         / |                           | \
|        /  |                           |  \
|    A* /___|___________________________|___\ B*
|           A    classical/on-shell     B
|
|
|------------------------------------------------> t


Fig. 4. Six points.

Since the virtual path $A^{*}\to A^{\prime}\to B^{\prime}\to B^{*}$ is an infinitesimal variation of the classical path $A^{*}\to A\to B\to B^{*}$, we conclude that the difference

$$S(A^{*}\to A^{\prime})+I(A^{\prime}\to B^{\prime})+S(B^{\prime}\to B^{*})$$ $$-S(A^{*}\to A)-S(A\to B)-S(B\to B^{*})$$ $$\tag{23} ~=~I(A^*\to A^{\prime}\to B^{\prime}\to B^*) -S(A^*\to A\to B\to B^*)~=~{\cal O}(\varepsilon^2)$$

cannot contain contributions linear in $\varepsilon$.

We next apply the Lemma and Corollary from Section 4. The six infinitesimal classical paths mentioned so far are all described by the differential (10), which is linear and hence obeys a (co-)vector addition rule. Therefore

$$\tag{24} S(A^{*}\to A^{\prime})-S(A^{*}\to A) +{\cal O}(\varepsilon^2) ~\stackrel{(10)}{=}~S(A\to A^{\prime}) ~\stackrel{(19)}{=}~\varepsilon Q_i,\qquad$$ $$\tag{25} S(B\to B^{*}) - S(B^{\prime}\to B^{*})+{\cal O}(\varepsilon^2) ~\stackrel{(10)}{=}~S(B\to B^{\prime}) ~\stackrel{(19)}{=}~\varepsilon Q_f.\qquad$$

Comparing eqs. (21)-(25), we arrive at the main conclusion of Noether's theorem, namely that the Noether charge is conserved,

$$\tag{26} Q_f~=~Q_i.$$

References:

1. R.P. Feynman, The Character of Physical Law, 1965, pp. 103 - 105.

--

$^1$ For Feynman's proof, see approximately 50 minutes into this video. Noether's theorem is covered in 45:25-51:27.

$^2$ Feynman uses the opposite convention for horizontal and vertical than this answer.

$^3$ Noether's theorem works with less assumptions, but to avoid mathematical technicalities, we impose assumption 1, 2 and 3. Note that it is easy to find examples that satisfies assumption 1 and 2, but where the classical path $q_{\rm cl}$ may jump discontinuously for varying boundary values $(q_f,t_f;q_i,t_i)$, so that assumption 3 is not satisfied.

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Qmechanic
answered Jan 24, 2012 by (3,120 points)
This is excellent, thanks, but I have a question: You say the virtual path $A^* -> A' -> B' -> B^*$ is an infinitesimal variation of the classical path $A^* -> A -> B -> B^*$, but the derivatives of the paths are not close at all. My understanding is that the Euler-Lagrange equations are only necessary for a weak extrema - i.e. the functional is stationary relative to all the curves 'close' to it, where we define 'close' in terms of both the value of $y$ and it's derivative $\dot{y}(t)$. But doesn't your proof assume that we have a strong extrema?

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user tom
Hi @tom. Thanks for the feedback. If you are asking for a mathematical rigorous treatment of variational/ functional derivatives (FD), you may want to ask on Mathematics or MathOverflow instead. Usually in (well-posed) physics problems, the existence and convergence of FD is not a problem. (If it is, it often signals ill-posed boundary conditions.) Assumptions 2 and 3 are in practice more critical.

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Qmechanic
@Qmechanic: I expanded my answer to include the simplest formal equivalent of Feyman's argument, and now I see that I ended up unintentionally plagiarizing your answer a little bit. I think I emphasized sufficiently different things to make my answer useful, and since your answer also overlaps my answer in a similar way, I hope you are not annoyed at this.

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Ron Maimon
+ 0 like - 1 dislike

Paths AB and CD are exactly the same length, with exactly the same start and end times, which means paths AC and BD are traversed in $\delta t = 0$, at an infinite velocity!

Mr Feynman: you're talking baloney, as you would say to a social scientist!

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Physiks lover
answered Feb 6, 2014 by (-20 points)
The "horizontal line" means perturbing the velocity from $\dot{x}$ to $\dot{x} + \epsilon \delta(t-t_0)$, where the perturbation is thought of as an infinitesimal kick at time $t_0$. This is not mathematically sensible by itself without thinking a bit about regulating the delta-function, but when you do regulate everything and cross the t's and dot the i's, Feynman's proof goes through and produces the shortest niftiest proof of Noether's theorem. It is nowadays standard to use a continuously varying kick $\dot{x} + \epsilon(t)$ instead, to avoid the limits-talk. See my answer.

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Ron Maimon
This is wrong, look at Ron's answer

This post imported from StackExchange Physics at 2015-05-03 09:39 (UTC), posted by SE-user Larry Harson

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