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$^3$:
that the Lagrangian $L(q,v,t)$ is a smooth function of its arguments $q$, $v$, and $t$.
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.
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:
- 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
Q&A (4870)
