Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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:

enter image description here

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?

share|improve this question
4  
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. –  Qmechanic Jan 23 '12 at 0:39
    
@Qmechanic Thanks. Google books didn't have an electronic copy of this book to link to. –  Mark Eichenlaub Jan 23 '12 at 0:40
1  
awesome question! –  Physiks lover Feb 6 at 15:42

3 Answers 3

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.

share|improve this answer

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.

Using horizontal lines is a little misleading, because when you make the argument precise, the lines are only infinitesimally different in velocity, 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 idea is as Feynman describes.

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.

share|improve this answer
3  
I get lost at the action of the little segment at the beginning. $S$ is a functional, not a function of $x, \dot{x}$, so I don't really understand what $\frac{\partial S}{\partial \dot{x}}$ means. Maybe it was supposed to be $\frac{\partial L}{\partial \dot{x}} \frac{\delta}{\epsilon}$, but I don't see how we get that, either. The action is just $S = \int L dt$ so it seems the action along the little path should be $\epsilon L(x,\frac{\delta}{\epsilon})$. What am I missing? –  Mark Eichenlaub Mar 23 '13 at 9:29
1  
-1: for stationary action, the start and end times remain fixed, and Feynman requires the distance travelled along paths AB and CD to be exactly the same, which means the velocites along AC and BD can't be finite –  Larry Harson Feb 5 at 23:52

Paths AB and CD are exactly the same length, with exactly the same start and end times, which means lengths AC and BD must be zero.

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.