I'm trying to understand the proof of Noether's Theorem in my Classical Mechanics class. We formulated it as follows:

A continous symmetry is defined as a flow $\phi^{\lambda}(q(t))$ which leaves the Lagrangian invariant, i.e.

$$L(\phi^{\lambda}(q(t)), \frac{\partial}{\partial t}\phi^{\lambda}(q(t)),t) = L(q(t),\dot{q}(t),t)$$

for each $\lambda$ and each $q(t)$ which solves the Euler-Lagrange equations.

The conserved quantity then is

$$\langle p,v(q)\rangle = \sum_{\alpha =1}^f p_{\alpha} v^{\alpha}(q)$$

where $p_{\alpha} = \frac{\partial L}{\partial \dot{q}^{\alpha}}$ is the conjugated momentum to $q^{\alpha}$ and $v(q)$ is the vector field implied by the flow $\phi^{\lambda}(q(t))$.

So the proof goes as follows:

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t} \Biggl[\sum_{\alpha =1}^f p_{\alpha} v^{\alpha}(q)\Biggr] &= \sum_{\alpha =1}^f \biggl[\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}^{\alpha}} \biggr] v^{\alpha}(q) + \sum_{\alpha =1}^f \frac{\partial L}{\partial \dot{q}^{\alpha}} \frac{\partial}{\partial t}v^{\alpha}(q) \\ &= \sum_{\alpha =1}^f \frac{\partial L}{\partial q^{\alpha}} \frac{\partial}{\partial \lambda} \phi^{\lambda}\bigl(q(t)\bigr) + \sum_{\alpha =1}^f \frac{\partial L}{\partial \dot{q}^{\alpha}} \frac{\partial}{\partial \lambda}\frac{\partial}{\partial t}\phi^{\lambda}\bigl(q(t)\bigr) \\ &= \frac{\mathrm{d}}{\mathrm{d}\lambda} L\biggl(\phi^{\lambda}\bigl(q(t)\bigr), \frac{\partial}{\partial t} \phi^{\lambda}\bigl(q(t)\bigr),t\biggr) \\ &= 0 \end{align}$$

where we used the Euler-Lagrange equations in the second line and the definition of the vector field $v(q)$. In the third line we used the definition of a continuous symmetry.

There's just one small thing that I don't fully understand now: If we calculate

$$\frac{\mathrm{d}}{\mathrm{d}\lambda} L\biggl(\phi^{\lambda}\bigl(q(t)\bigr), \frac{\partial}{\partial t} \phi^{\lambda}\bigl(q(t)\bigr),t\biggr) = \frac{\partial L}{\partial \phi^{\lambda}}\frac{\partial}{\partial \lambda} \phi^{\lambda} + \frac{\partial L}{\partial (\frac{\partial}{\partial t} \phi^{\lambda} )} \frac{\partial}{\partial \lambda} \frac{\partial }{\partial t} \phi^{\lambda}$$

(I dropped the summation here). Why is the derivative towards $\phi^{\lambda}$ immediately the same as the derivative towards $q(t)$? I know the flow $\phi^{\lambda}$ maps physical solutions to physical solutions but I don't quite intuitively see how the two derivatives are the same. I hope someone can clear this up for me!

  • $\begingroup$ in both cases, it's the partial derivative of $L$ with respect to its first variable; it might be less confusing if you'd given the variable and the trajectory a different name, eg $L=L(x,v,t)$ and use $\partial L/\partial x$, regardless if you evaluate the function at $(q(t),\dot q(t), t)$ or $(\phi^\lambda(q(t)),\partial_t\phi^\lambda(q(t)),t)$ $\endgroup$ – Christoph Jun 9 '13 at 10:29
  • $\begingroup$ @Christoph But why does choosing a different variable for the trajectory that depends on the old variable not change the function? If $\phi^{\lambda}$ where no function of $q(t)$ I'd see what you mean, but if we express our Langrangian through a new variable which is a function of the old variables, the Lagrangian should change, and hence the derivative should too. What am I missing? $\endgroup$ – user17574 Jun 9 '13 at 10:46
  • $\begingroup$ it's the chain rule - regardless if you're looking at $L$, $L\circ q$ or $L\circ\phi^\lambda\circ q$, the leftmost term will always be $\partial L/\partial x$ $\endgroup$ – Christoph Jun 9 '13 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy