Consider a coordinate system in which the point A has coordinate $z_0$ and point B has ccordinate $z$ where there is a particle of mass $m$. Let there be a constant gravitational field of Earth. The Lagrangian of a particle in this system is given by $$L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)-mg(z-z_0)$$ where the potential energy of point B is measured w.r.t point A. Now consider different coordinate system where the origin is shifted from $x=y=z=0$ to $x=y=0,z=\alpha$. In this coordinate, A has coordinate $z_0+\alpha$ and B has coordinate $z+\alpha$. Therefore, although the coordinate $z$ is not cyclic, the Lagrangian remains invariant under translation in the $z$ direction. Therefore, by Noether's theorem, the momentum $p_z$ should be conserved. But as we know this is not the case due to constant gravitational field along $z$-direction. Is this not a contradiction?

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    $\begingroup$ If your $z_0$ is a fixed point, then it is not acted upon by translation. By definition, a symmetry transformation can only act on the dynamical variables, not on constants, so your Lagrangian is not invariant. $\endgroup$ – ACuriousMind Aug 2 '16 at 17:19
  • $\begingroup$ A is a fixed point. $z_0$ is the coordinate label of A that will change under the change of coordinates system. I don't get your point. Think of a coordinate system and translating it. $\endgroup$ – SRS Aug 2 '16 at 17:26
  • $\begingroup$ Here I'm talking about translating the coordinate system not the particle itself. If I changed the location of the particle B, then yes, you're right. $\endgroup$ – SRS Aug 2 '16 at 17:30
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    $\begingroup$ You have to look at what a transformation for the Noether theorem is - it is a transformation of the dynamical variables. Your translation of the coordinate system does not fulfill the technical prerequisites of Noether's theorem. $\endgroup$ – ACuriousMind Aug 2 '16 at 17:37
  • $\begingroup$ Okay. Now this makes sense to me. It is true that if a coordinate is cyclic, the corresponding linear momentum is conserved (as can be seen from Euler-Lagrange equation). Is the reverse also true? I mean, can the Lagrangian be invariant under $q_i\rightarrow q_i+\alpha$ without $q_i's$ being cyclic? Or should I post it as a different question? $\endgroup$ – SRS Aug 2 '16 at 17:48
  1. ACuriousMind has already pointed out in comments the crucial point: $z_0$ is an external parameter, not an dynamical variable of the action $S[x,y,z]$.

  2. OP then ponders in a comment what happens if we artificially promote $z_0$ to a dynamical variable of the action $S[x,y,z,z_0]$? Nevermind this possibly does not make physical sense. What would be the mathematical consequences for Noether's theorem?

  3. OK, let's investigate. The translations $$ z ~\longrightarrow ~ z+\alpha, \qquad z_0 ~\longrightarrow ~ z_0+\alpha $$ is indeed an exact off-shell symmetry of the action $S[x,y,z,z_0]$. Noether's theorem then therefore states that the corresponding Noether charge $$p_z~:=~\frac{\partial L}{\partial \dot{z}}$$ is conserved on-shell $$ \frac{dp_z}{dt}~\approx~0. $$ [Here the $\approx$ symbol means equality modulo eom.]

  4. Only trouble is that the eom for $z_0$ reads $$ 0~\approx~\frac{\delta S}{\delta z_0}~=~mg, $$ i.e. if we want to put the system on-shell, we would have to turn off gravity! And then $p_z$ would be conserved on-shell. Therefore Noether's theorem is not violated.


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