2
$\begingroup$

So as we all know for a system that has translational symmetry Noether's Theorem states that momentum is conserved, more precisely the theorem states that the quantity: $$\frac{\partial L}{\partial \dot{q}}$$ so the generalized momentum is conserved. Here i have a problem: suppose i want to show that classical momentum $p=mv$ is conserved in a system with traslational symmetry (also of course potential energy in the Lagrangian does not depend on velocity) I then have: $$\frac{\partial L}{\partial \dot{x}}=\frac{\partial K}{\partial \dot{x}}=\frac{\partial}{\partial \dot{x}}\frac{1}{2}m\dot{x}^2=m\dot{x}.$$ Perfect! But suppose that i want to use a parametrization for my system, so: $$x(t)=\Gamma(q(t))$$ as we usually do in Lagrangian Mechanics, then I have that the conserved quantity is still: $$\frac{\partial L}{\partial \dot{q}}.$$ In fact Noether's Theorem states that generalised momentum is conserved and this is by definition the generalized momentum. Well then I have: $$\frac{\partial L}{\partial \dot{q}}=\frac{\partial}{\partial \dot{q}}\frac{1}{2}m\dot{q}^2|\Gamma ' (q)|^2=m\dot{q}|\Gamma ' (q)|^2=mv|\Gamma ' (q)|.$$ WTF is this?? Furthermore if i choose $\Gamma$ to represent a line with the following parametrization: $$\Gamma = \begin{bmatrix}kq \\ 0 \\ 0\end{bmatrix}.$$ I get: $$\frac{\partial L}{\partial \dot{q}}=mv|k|$$ so the conserved quantity depends on the parametrization??? Now: I know of course that I made a mistake somewhere; maybe on the content of Noether's Theorem (even if i took the content of said theorem straight from my book of Lagrangian Mechanics) or maybe somewhere else. My questions are:

  1. Why I get this result?
  2. How can i show that momentum $p=mv$ is conserved for a symmetrically translational system using Noether's Theorem and using any parametrization $\Gamma$ I want?
  3. Is it true that generalised momentum is conserved for any symmetrically translational system?
  4. When conservation of generalised momentum implies conservation of classical momentum?

This is my problem; hope you can help me out. Please try to give me a complete answer, this problem is bugging me a lot.

$\endgroup$
  • $\begingroup$ Is $\Gamma$ a function of $\dot{q} $. $\endgroup$ – Blaze May 28 at 16:20
3
$\begingroup$
  1. Let us for simplicity consider a 1D system. If the Lagrangian $L(\dot{x},t)$ has a cyclic variable $x$, then the action has an infinitesimal translation symmetry $$\delta x~=~\epsilon,$$ and it is well-known that the conserved Noether charge $$ Q~=~\frac{\partial L}{\partial \dot{x}}\tag{1} $$ is the conjugate momentum.

  2. OP considers next a coordinate transformation $$x~=~f(q,t).$$ Note that $q$ is not necessarily a cyclic variable (because $\dot{x}=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial t}$ may depend on $q$). The new symmetry becomes $$ \delta q~=~\epsilon Y,$$ where $$Y~=~\frac{\partial q}{\partial x}~=~\left(\frac{\partial f}{\partial q}\right)^{-1}$$ is the so-called generator. According to Noether's formula, the conserved Noether charge is "momentum times generator": $$ Q~=~\frac{\partial L}{\partial \dot{q}} Y~=~\frac{\partial L}{\partial \dot{x}},\tag{2}$$ which is the same as before because of the chain rule.

| cite | improve this answer | |
$\endgroup$

Your Answer

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

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